Record
Recursive sequence矩阵快速幂
#include <bits/stdc++.h>
#include<stdint.h>
#define int long long
#define scan(n) scanf("%lld", &(n))
#define scann(n, m) scanf("%lld%lld", &(n), &(m))
#define scannn(a, b, c) scanf("%lld%lld%lld", &(a), &(b), &(c))
#define pb push_back
#define fo(i, a, b) for (int i = (a); i <= (b); i++)
const int N=8;//N个系数，N维矩阵
using namespace std;
const int mod=2147493647;
struct matrix{int m[10][10];};
matrix ans,base,m;
int n,a,b;
matrix multi(matrix a, matrix b){
matrix tmp;
for(int i=1;i<=N;++i){
for (int j=1;j<=N;++j){
tmp.m[i][j]=0;
for(int k=1;k<=N;++k)
tmp.m[i][j]=(tmp.m[i][j]+(a.m[i][k]*b.m[k][j]))%mod;
}
}
return tmp;
}
matrix fastm(matrix base, int n){
matrix ans;
memset(ans.m,0,sizeof(ans.m));
for (int i=1;i<=N;i++)ans.m[i][i]=1;
while(n){
if(n&1)ans=multi(ans,base);
base=multi(base, base);
n>>=1;
}
return ans;
}
int32_t main(){
matrix m,ans;
memset(m.m,0,sizeof(m.m));
m.m[1][1]=1;m.m[1][2]=2;m.m[1][4]=1;m.m[1][5]=4;m.m[1][6]=6;m.m[1][7]=4;m.m[1][8]=1;
m.m[2][1]=1;
m.m[3][2]=1;
m.m[4][4]=1;m.m[4][5]=4;m.m[4][6]=6;m.m[4][7]=4;m.m[4][8]=1;
m.m[5][5]=1;m.m[5][6]=3;m.m[5][7]=3;m.m[5][8]=1;
m.m[6][6]=1;m.m[6][7]=2;m.m[6][8]=1;
m.m[7][7]=1;m.m[7][8]=1;
m.m[8][8]=1;
int T;scan(T);
while(T--){
scannn(n,a,b);
ans = fastm(m,n-3);
int f[10],x=0;
f[3]=a,f[2]=b,f[1]=(2*a+b+3*3*3*3),f[4]=3*3*3*3,f[5]=3*3*3,f[6]=3*3,f[7]=3,f[8]=1;
fo(i,1,8){
x=x%mod+((f[i]%mod)*(ans.m[1][i])%mod)%mod;
x%=mod;
}
printf("%lld\n",x);
}
return 0;
矩阵快速幂，用于求解递推式。
Happy Necklace 矩阵快速幂
注意状态的转移，很有意思~
#include <bits/stdc++.h>
#include<stdint.h>
using namespace std;
#define int long long
#define scan(n) scanf("%lld", &(n))
#define ms(a) memset(a, 0, sizeof(a))
#define fo(i, a, b) for (int i = (a); i <= (b); i++)
#define ro(i, a, b) for (int i = (a); i >= (b); i--)
#define dbg(args...) do {cout << #args << " : "<< args << endl;}
const int inf = 0x3f3f3f3f;
const int mod=1e9+7;
const int N=3;
struct matrix{
int m[N+1][N+1];
};
matrix e,d;
matrix qpow(matrix a,int b){
matrix ans=e;
while(b){
if(b&1){
matrix x=d;
for(int i=0;i<=N-1;i++)
for(int j=0;j<=N-1;j++)
for(int k=0;k<=N-1;k++)
x.m[i][j]=x.m[i][j]%mod+ans.m[i][k]*a.m[k][j]%mod;
ans=x;
}
matrix x=d;
for(int i=0;i<=N-1;i++)
for(int j=0;j<=N-1;j++)
for(int k=0;k<=N-1;k++)
x.m[i][j]=x.m[i][j]%mod+a.m[i][k]*a.m[k][j]%mod;
a=x;
b>>=1;
}
return ans;
}
int32_t main(){
int T;scan(T);
matrix base,m;
for(int i=0;i<=N-1;i++){
for(int j=0;j<=N-1;j++){
base.m[i][j]=0;
m.m[i][j]=0;
e.m[i][j]=0;
d.m[i][j]=0;
}
}
base.m[0][0]=1;
base.m[1][0]=1;
base.m[2][0]=1;
m.m[0][0]=1;m.m[0][1]=0;m.m[0][2]=1;
m.m[1][0]=1;m.m[1][1]=0;m.m[1][2]=0;
m.m[2][0]=0;m.m[2][1]=1;m.m[2][2]=0;
for(int i=0;i<=N-1;i++)e.m[i][i]=1;
while(T--){
int n;scan(n);
int ans=0;
if(n==1)ans=2;
else if(n==2)ans=3;
else{
matrix temp=qpow(m,n-2);
matrix x=d;
for(int i=0;i<=N-1;i++)
for(int j=0;j<=N-1;j++)
for(int k=0;k<=N-1;k++)
x.m[i][j]=x.m[i][j]%mod+temp.m[i][k]*base.m[k][j]%mod;
temp=x;
ans=temp.m[0][0]+temp.m[1][0]+temp.m[2][0];
ans%=mod;
}
printf("%lld\n",ans);
}
}
Counting cliques DFS
include <bits/stdc++.h>
include<stdint.h>
using namespace std;
define int long long
define scan(n) scanf("%lld", &(n))
define scann(n, m) scanf("%lld%lld", &(n), &(m))
define scannn(a, b, c) scanf("%lld%lld%lld", &(a), &(b), &(c))
define pb push_back
define fo(i, a, b) for (int i = (a); i <= (b); i++)
const int inf = 0x3f3f3f3f;
const int maxn = 2e5+100;
int n,m,s,x,y,pre[20],e120,ans,tot,head[120];
struct node{
int v,nxt;
}a[1024];
void add(int u,int v){
tot++;
a[tot].v=v;
a[tot].nxt=head[u];
head[u]=tot;
}
void dfs(int u,int step){
pre[step]=u;
if(step==s){ans++;return;}
for(int j=head[u];~j;j=a[j].nxt){
int x=a[j].v;
int f=1;
fo(i,1,step){
if(epre[i]==0){f=0;break;}
}
if(f){
step++;
dfs(x,step);
step--;
}
}
return;
}
int32_t main() {
int T;scan(T);
while(T--){
scannn(n,m,s);
fo(i,0,15)pre[i]=0;
fo(i,0,105)fo(j,0,105)ei=0;
fo(i,0,105)head[i]=-1;
ans=0,tot=0;
fo(i,1,m){
scann(x,y);
if(x>y)swap(x,y);
ex=1;add(x,y);
}
fo(i,1,n)dfs(i,1);
printf("%lld\n",ans);
}
return 0;
}
Distance in Tree
Tree painting 树形DP
#include <bits/stdc++.h>
#include<stdint.h>
#define int long long
#define scan(n) scanf("%lld", &(n))
#define scann(n, m) scanf("%lld%lld", &(n), &(m))
#define scannn(a, b, c) scanf("%lld%lld%lld", &(a), &(b), &(c))
#define prin(n) printf("%lld", (n))
#define pb push_back
#define mp make_pair
#define ms(a) memset(a, 0, sizeof(a))
#define fo(i, a, b) for (int i = (a); i <= (b); i++)
#define ro(i, a, b) for (int i = (a); i >= (b); i--)
const int inf = 0x3f3f3f3f;
using namespace std;
const int maxn = 2e5+100;
vector<int>e[maxn];
int siz[maxn];
int ans,n;
int u,v;
void getsize(int v,int u){
siz[v]=1;
for(int x:e[v]){
if(x==u)continue;
getsize(x,v);
siz[v]+=siz[x];
}
}
void dfs(int v,int u,int t){
ans=max(ans,t);
cout<<v<<" "<<u<<" "<<t<<endl;
for(int x:e[v]){
if(x==u)continue;
dfs(x,v,t-2*siz[x]+n);
}
}
int32_t main() {
scan(n);
fo(i,1,n-1){
scann(u,v);
e[u].pb(v);e[v].pb(u);
}
getsize(1,0);
int t=0;
fo(i,1,n)t+=siz[i];
dfs(1,0,t);
printf("%lld",ans);
return 0;
}
ABOUT 笛卡尔树
单调栈
http://www.hankcs.com/program/algorithm/poj-3250-bad-hair-day.html
https://blog.csdn.net/wubaizhe/article/details/70136174
https://ac.nowcoder.com/acm/contest/view-submission?submissionId=40898761
https://blog.csdn.net/u013554860/article/details/51360542
https://xlor.cn/2019/4/2019nanchang/
https://blog.csdn.net/lr7682/article/details/90247997
牛客第一场AEquivalent Prefixes-前缀笛卡尔树
序列u，v 对于\([1,ans]\)上所有的\([L,R]\)\((1<=L<=R<=ans<=n)\)
都满足\(RMQ(u,L,R)=RMQ(v,L,R)\)
求\(max(ans)\);
分析：考虑判断两个序列的前缀笛卡尔树是否相等
注意，如果前缀笛卡尔树相等，则可以判断每一个栈深都相同，想想为什么？、
#include <bits/stdc++.h>
#include<stdint.h>
#define int long long
#define scan(n) scanf("%lld", &(n))
#define scann(n, m) scanf("%lld%lld", &(n), &(m))
#define scannn(a, b, c) scanf("%lld%lld%lld", &(a), &(b), &(c))
#define prin(n) printf("%lld", (n))
#define pb push_back
#define mp make_pair
#define ms(a) memset(a, 0, sizeof(a))
#define fo(i, a, b) for (int i = (a); i <= (b); i++)
#define ro(i, a, b) for (int i = (a); i >= (b); i--)
const int inf = 0x3f3f3f3f;
using namespace std;
const int maxn = 1e5+100;
int n;
int a[maxn],b[maxn];
int s1[maxn],s2[maxn];
int32_t main() {
while(scan(n)!=EOF){
fo(i,1,n)scan(a[i]);
fo(i,1,n)scan(b[i]);
int head1=1,head2=1,tail1=1,tail2=1;
s1[head1]=s2[head2]=inf;
int maxpos=1;
fo(i,1,n){
while(head1<=tail1&&s1[tail1]>a[i])tail1--;
while(head2<=tail2&&s2[tail2]>b[i])tail2--;
//cout<<"before:\n";
//cout<<"tail1: "<<tail1<<" s1[tail1]:"<<s1[tail1]<<endl;
//cout<<"tail2: "<<tail2<<" s2[tail2]:"<<s2[tail2]<<endl;
if(tail1!=tail2)break;
s1[++tail1]=a[i],s2[++tail2]=b[i];
//cout<<"after:\n";
//cout<<"tail1: "<<tail1<<" s1[tail1]:"<<s1[tail1]<<endl;
//cout<<"tail2: "<<tail2<<" s2[tail2]:"<<s2[tail2]<<endl;
maxpos=i;
}
cout<<maxpos<<endl;
}
return 0;
}
笛卡尔树习题
http://acm.hdu.edu.cn/showproblem.php?pid=6305
https://blog.csdn.net/zhaiqiming2010/article/details/80245872
CF1199FRectangle Painting 1-区间DP
给定一个\(n \times n\)的矩阵,由\("\#”,"."\)组成，每次可以将\(h\times w\)矩形内的\("\#"\)变成\("."\)
花费为\(max(h,w)\),求将所有\("\#"\)变成\("."\)的最小花费
#include<bits/stdc++.h>
#define fo(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
char a[55][55];
int f[55][55][55][55];
int dp(int x,int y,int xx,int yy){
if(f[x][y][xx][yy]!=-1)return f[x][y][xx][yy];
int ans=max(xx-x+1,yy-y+1);
fo(i,x,xx-1)ans=min(ans,dp(x,y,i,yy)+dp(i+1,y,xx,yy));
fo(i,y,yy-1)ans=min(ans,dp(x,y,xx,i)+dp(x,i+1,xx,yy));
return f[x][y][xx][yy]=ans;
}
int main(){
int n;scanf("%lld",&n);
fo(i,1,n)scanf("%s",a[i]+1);
memset(f,-1,sizeof(f));
fo(x,1,n){
fo(y,1,n){
fo(xx,1,n){
fo(yy,1,n){
if(x>xx||y>yy)f[x][y][xx][yy]=0;
if(x==xx&&y==yy)f[x][y][xx][yy]=(a[x][y]=='#'?1:0);
}
}
}
}
cout<<dp(1,1,n,n)<<endl;
}
CF1190EMatching vs Independent Set 图匹配
给定一个3*n个点m条边的简单无向图，要求在这个图里找出一个有n条边的连通图
或者找出一个有n个点的独立点集，并且输出答案
#include <bits/stdc++.h>
#include<stdint.h>
#define int long long
#define scan(n) scanf("%lld", &(n))
#define scann(n, m) scanf("%lld%lld", &(n), &(m))
#define scannn(a, b, c) scanf("%lld%lld%lld", &(a), &(b), &(c))
#define prin(n) printf("%lld", (n))
#define pb push_back
#define mp make_pair
#define ms(a) memset(a, 0, sizeof(a))
#define fo(i, a, b) for (int i = (a); i <= (b); i++)
#define ro(i, a, b) for (int i = (a); i >= (b); i--)
const int inf = 0x3f3f3f3f;
using namespace std;
const int maxn = 3e5+100;
int n,m;
int u[maxn],v[maxn];
int vis[maxn];
int32_t main() {
int T;scan(T);
while(T--){
scann(n,m);
fo(i,1,3*n)vis[i]=0;
vector<int>edge_ans;
fo(i,1,m){
int u,v;scann(u,v);
if(vis[u]||vis[v])continue;
vis[u]=1,vis[v]=1;
edge_ans.pb(i);
}
if(edge_ans.size()>=n){
printf("Matching\n");
fo(i,0,n-1){
printf("%lld ",edge_ans[i]);
}
}
else{
printf("IndSet\n");
int cnt=0;
fo(i,1,3*n){
if(cnt>=n)break;
if(!vis[i])printf("%lld ",i),cnt++;
}
}
printf("\n");
}
return 0;
}
盒子-组合数
隔板法；
子问题：给定n个物品，将它们分成m堆，保证每堆的物品个数不少于h个
则有\(C_{n-m*h-1}^{m-1}\)种分割方法
#include <bits/stdc++.h>
#include<stdint.h>
using namespace std;
#define int long long
#define scan(n) scanf("%lld", &(n))
#define scann(n, m) scanf("%lld%lld", &(n), &(m))
#define scannn(a, b, c) scanf("%lld%lld%lld", &(a), &(b), &(c))
#define prin(n) printf("%lld", (n))
#define pb push_back
#define mp make_pair
#define ms(a) memset(a, 0, sizeof(a))
#define fo(i, a, b) for (int i = (a); i <= (b); i++)
#define ro(i, a, b) for (int i = (a); i >= (b); i--)
#define dbg(args...) do {cout << #args << " : "<< args << endl;}
const int inf = 0x3f3f3f3f;
const int maxn = 1e5+100;
const int mod=1e9+7;
int f[maxn],inv[maxn];
int qpow(int a,int b){
int res=1;
while(b){
if(b&1)res=res*a%mod;
a=a*a%mod;;
b/=2;
}
return res;
}
void init(){
f[0]=1;
for(int i=1;i<=maxn-1;i++)f[i]=f[i-1]*i%mod;
inv[maxn-1]=qpow(f[maxn-1],mod-2);
for(int i=maxn-2;i>=0;i--)inv[i]=inv[i+1]*(i+1)%mod;
}
int C(int n,int m){
return f[n]*inv[m]%mod*inv[n-m]%mod;
}
int32_t main() {
int f,w,h;
cin>>f>>w>>h;
int fm=0,fz=0;
init();
for(int i=1;i<=w;i++){
fm+=(C(w-1,i-1)*C(f+1,i))%mod;
fm%=mod;
}
for(int i=1;i<=w/(h+1);i++){
fz=fz+C(w-i*h-1,i-1)*C(f+1,i)%mod;
fz%=mod;
}
int ans=fz*qpow(fm,mod-2)%mod;
cout<<ans<<endl;
return 0;
}
组合数部分：
for(int i=1;i<=w;i++){
fm=fm+(C(w-1,i-1)*C(f-1,i-1))*2%mod;
fm=fm+(C(w-1,i-1)*C(f-1,i-2))%mod;
fm=fm+(C(w-1,i-1)*C(f-1,i))%mod;
fm%=mod;
}
for(int i=1;i<=w/(h+1);i++){
fz=fz+C(w-i*h-1,i-1)*C(f-1,i-1)*2%mod;
fz=fz+C(w-i*h-1,i-1)*C(f-1,i-2)%mod;
fz=fz+C(w-i*h-1,i-1)*C(f-1,i)%mod;
fz%=mod;
}
可以合并为：
for(int i=1;i<=w;i++){
fm+=(C(w-1,i-1)*C(f+1,i))%mod;
fm%=mod;
}
for(int i=1;i<=w/(h+1);i++){
fz=fz+C(w-i*h-1,i-1)*C(f+1,i)%mod;
fz%=mod;
}
找朋友-状压DP
赵队出的题，找朋友
#include <bits/stdc++.h>
#include<stdint.h>
using namespace std;
#define int long long
#define scan(n) scanf("%lld", &(n))
#define scann(n, m) scanf("%lld%lld", &(n), &(m))
#define scannn(a, b, c) scanf("%lld%lld%lld", &(a), &(b), &(c))
#define prin(n) printf("%lld", (n))
#define pb push_back
#define mp make_pair
#define ms(a) memset(a, 0, sizeof(a))
#define fo(i, a, b) for (int i = (a); i <= (b); i++)
#define ro(i, a, b) for (int i = (a); i >= (b); i--)
#define dbg(args...) do {cout << #args << " : "<< args << endl;}
const int inf = 0x3f3f3f3f;
const int maxn = (1<<21);
int x[25],y[25];
int n,f;
double dp[maxn];
bool check(int s,int t){
for(int i=1;i<=n;i++){
int x=(1<<i);
if((!(s&x))&&(t&x))return false;
}
return true;
}
double dis(int i,int j){
return sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]));
}
double min_dis(int t,int j){
vector<double>v;
for(int i=1;i<=n;i++){
int x=(1<<i);
if(t&x){
v.pb(dis(i,j));
}
}
sort(v.begin(),v.end());
if(v.size()<=1)return inf;
return v[0]+v[1];
}
int32_t main() {
int T;scan(T);
while(T--){
scan(n);
int s=0;
int xx=((1<<(n+1))-1)^1;
for(int i=0;i<=maxn-1;i++)dp[i]=inf;
fo(i,1,n){
scannn(x[i],y[i],f);
if(f)s^=(1<<i);
}
dp[s]=0;
for(int i=s;i<=xx;i++){
if(!check(i,s))continue;
fo(j,1,n){
int t=(1<<j);
if((i&t)&&(!(s&t))){
int x=i^t;
dp[i]=min(dp[x]+min_dis(x,j),dp[i]);
}
else continue;
}
}
if(dp[xx]==inf){
cout<<"No Solution\n";
continue;
}
printf("%.6lf\n",dp[xx]);
}
return 0;
}
找质数-素数筛
因为数字范围\(<=1e14\),所以在快速幂的过程中会爆long long，要用__int128
也可以结合费马小定理 \(a^{p-1} \ mod \ p == 1\),P为质数，已知质数间隔小于200
#include <bits/stdc++.h>
#include<stdint.h>
using namespace std;
const int maxn=1e7;
__int128 read(){
__int128 x=0,f=1;
char ch=getchar();
while(ch<'0'||ch>'9'){
if(ch=='-')f=-1;
ch=getchar();
}
while(ch>='0'&&ch<='9'){
x=x*10+ch-'0';
ch=getchar();
}
return x*f;
}
void print(__int128 x){/*__int128输出*/
if(x<0)putchar('-'),x=-x;
if(x>9)print(x/10);
putchar(x%10+'0');
}
struct Prime{
int prime_cnt;
int prime[maxn+5],vis[maxn+5];
int p[maxn+5];
void Prime_init(){/*将合数标记为vis[i]=1*/
prime_cnt=0;
for(int i=2;i<=maxn;i++){
if(!vis[i])prime[++prime[0]]=i;
for(int j=1;j<=prime[0]&&i*prime[j]<=maxn;j++){
vis[i*prime[j]]=1;
if(i%prime[j]==0)break;
}
}
/*prime_cnt记录素数的数量*//*p数字记录每个素数*/
for(int i=2;i<maxn;i++)if(!vis[i])p[++prime_cnt]=i;
}
bool isprime(__int128 x){
for(int i=1;i<=prime_cnt;i++){
if(p[i]>=x)return true;
if(x%p[i]==0)return false;
}
return true;
}
}F;
__int128 qpow(__int128 a,__int128 b,__int128 mod){/*快速幂*/
__int128 res=1;
while(b){
if(b&1)res=res*a%mod;
a=a*a%mod;
b>>=1;
}
return res;
}
int32_t main() {
int T;cin>>T;
F.Prime_init();
while(T--){
__int128 a,p;p=read(),a=read();
for(__int128 q=p-1;q>=1;q--){
if(F.isprime(q)){
__int128 ans=qpow(a,q,p);
print(ans);
printf("\n");
break;
}
}
}
return 0;
}
Codeforces204E KMP
给定一主串,再查询某字符串是否可以由主串的两个子串拼接而成.
方法是顺着kmp一遍,再反着kmp一遍,记录查询的字符串的前缀和后缀在主串中的位置的最小值
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10005;
int z[maxn];
int pi[maxn];
int pos1[1005], pos2[1005];
int ans;
string s, t;
void prefix(string s) {
int n = s.size();
for (int i = 0; i < n;i++)
pi[i] = 0;
for (int i = 1; i < n; i++)
{
int j = pi[i - 1];
while (j > 0 && s[i] != s[j]) j = pi[j - 1];
if (s[i] == s[j]) j++;
pi[i] = j;
}
}
void kmp(string s,string t)//s是待匹配的
{
for (int i = 0; i <= 1000;i++)
pos1[i] = pos2[i] = maxn + 5;
int n = s.size();
int m = t.size();
string ss = s + "#" + t;
prefix(ss);
for (int i = n + 1; i <= m + n; i++)
{
int t = pi[i];
pos1[t] = min(pos1[t], i - n);
}
ss = s + "#" + t;
prefix(ss);
for (int i = n + 1; i <= m + n; i++)
{
int t = pi[i];
pos2[t] = min(pos2[t], i - n);
}
for (int i = 1; i <= n-1;i++)
{
int x = pos1[i];
int y = pos2[n - i];
if(x<=m-y){
ans++;
break;
}
}
}
int main()
{
cin >> t;
int q;
scanf("%d", &q);
for (int i = 0; i < q;i++)
{
cin >> s;
kmp(s, t);
}
cout << ans << endl;
}
Codeforces816E 树上背包
注意最后求答案的时候一定是\(dp[1][i][0]\)或者是\(dp[1][i][1]\)
因为想要使用优惠券,就必须保证它们的父亲被选了,而点1就是根节点
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int inf = 1e15;
const int maxn = 5000 + 10;
int n, b;
int c[maxn], d[maxn], x[maxn];
int dp[maxn][maxn][2];
int sz[maxn];
vector<int> G[maxn];
void dfs(int u)
{
sz[u] = 1;
dp[u][1][0] = c[u];
dp[u][1][1] = c[u] - d[u];
dp[u][0][0] = 0;
for(int v:G[u]){
dfs(v);
for (int i = sz[u]; i >= 0; i--)
{
for (int j = 0; j <= sz[v];j++)
{
dp[u][i + j][0] = min(dp[u][i + j][0], dp[u][i][0] + dp[v][j][0]);
dp[u][i + j][1] = min(dp[u][i + j][1], dp[u][i][1] + dp[v][j][0]);
dp[u][i + j][1] = min(dp[u][i + j][1], dp[u][i][1] + dp[v][j][1]);
}
}
sz[u] += sz[v];
}
}
int32_t main()
{
scanf("%lld%lld", &n, &b);
scanf("%lld%lld", &c[1], &d[1]);
for (int i = 2; i <= n;i++)
{
scanf("%lld%lld%lld", &c[i], &d[i], &x[i]);
G[x[i]].push_back(i);
}
for (int i = 0; i <= n;i++)
for (int j = 0; j <= n;j++)
dp[i][j][0] = dp[i][j][1] = inf;
dfs(1);
int ans = 0;
for (int j = 1; j <= n; j++)
if(dp[1][j][0]<=b||dp[1][j][1]<=b)
ans = max(ans, j);
cout << ans << endl;
}
NowCoder 非递归实现组合型枚举
模拟汇编的操作
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 100;
vector<int> v;
int n, m;
int st[maxn], address, top;
/*
void cal(int x){//参数x表示当前选择的数字
if(v.size()>m||v.size()+(n-x+1)<m)
return;
if(x==n+1){
for (int i = 0; i < v.size();i++){
printf("%d ", v[i]);
}
printf("\n");
return;
}
//选x
v.push_back(x);
cal(x + 1);
//不选x
v.pop_back();
cal(x + 1);
return;
}*/
void cal(int x,int ret_addr){//新指令(数字)进栈
int old_top = top;
st[++top] = x;
st[++top] = ret_addr;
st[++top] = old_top;
}
int ret(){
int ret_addr = st[top - 1];
top = st[top];//让最后一条语句出栈
return ret_addr;
}
int main(){
cin >> n >> m;
cal(1, 0);//数字1进栈
while(top){
int x = st[top - 2]; //获取的是数字x
switch(address){
case 0:
if(v.size()>m||v.size()+(n-x+1)<m){
address = ret(); //也就是return;
continue;
}
if(x==n+1){
for (int i = 0; i < v.size();i++){
printf("%d ", v[i]);
}
printf("\n");
address = ret();
continue;
}
v.push_back(x);
cal(x + 1, 1);
address = 0;
continue;
case 1:
v.pop_back();
cal(x + 1, 2);//返回后从cas2继续执行
address = 0;
continue;
case 2:
address = ret();
}
}
}
CCPC秦皇岛-KMP
把原串反过来...而已QuQ
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e7 + 5;
char s[maxn], t[maxn];
vector<int> prefix_function(int n){
vector<int> pi(n);
for (int i = 1; i < n;i++){
int j = pi[i - 1];
while(j>0&&t[i]!=t[j])j = pi[j - 1];
if(t[i]==t[j])j++;
pi[i] = j;
}
return pi;
}
int main(){
int a, b;
while(cin>>a>>b){
scanf("%s", s);
long long ans = 1e9 * 1e7 * 2;
ans = -ans;
int sz = strlen(s);
int l = 0;
for (int i = sz - 1; i >=0; i--)
{
if(s[i]=='.')break;
t[l++] = s[i];
}
vector<int> pi = prefix_function(l);
for (int i = 0; i < l; i++)
{
long long pos = i + 1;
long long len = i + 1;
len = pos - pi[i];
ans = max(ans, a * pos - b * len);
}
cout << ans << endl;
}
}
Tree-DP
void dp(int u)
{
sz[u] = 1;
init();
f[u][1] = val[x];
for(int v:to[u])
{
dp(v);
for (int i = sz[u];i>0;i--)
{
for (int j = 1; j <= sz[u];j++)
{
f[u][i + j] = max(f[u][i + j], f[u][i] + f[v][j]);
}
}
sz[u] += sz[v];
}
}
void dp(int u)
{
for(int v:to[u])
{
for (int i = 1; i <= m;i++)
f[v][i] = f[u][i];
dp(v);
for (int i = weight[v]; i <= m;i++)
f[u][i] = max(f[u][i], f[v][i - weight[v]] + value[v]);
}
}
点分治
树,n个点,边带权,求问边权和小于k的路径条数
对于指定的根节点p,树上的路径分为:
1.经过p (x ~ p)+(p ~ y )
2.包含于p的一棵子树,且不经过p(递归处理)
HDU2222 AC自动机
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 5;
struct Trie {
int nxt[maxn][26], fail[maxn], end[maxn];
int root ,L;
int newnode(){
for (int i = 0; i < 26; i++)
nxt[L][i] = -1;
end[L++] = 0;
return L - 1;//返回节点的数量
}
void init() {
L = 0;
root = newnode();
}
void insert(char buf[]) {
int len = strlen(buf);
int now = root;
for (int i = 0; i < len; i++)
{
int c = buf[i] - 'a';
if(nxt[now][c]==-1){
nxt[now][c] = newnode();
}
now = nxt[now][c];
}
end[now]++;
}
void build(){
queue<int> q;
fail[root] = root;
for(int i=0;i<26;i++) {
if (nxt[root][i] == -1) {
nxt[root][i] = root;//empty,指向根
}
else{
fail[nxt[root][i]] = root;
q.push(nxt[root][i]);
}
}
while(!q.empty()) {
int now = q.front();
q.pop();
for(int i=0;i<26;i++) {
if(nxt[now][i] == -1){
nxt[now][i] = nxt[fail[now]][i];
}
else{
fail[nxt[now][i]] = nxt[fail[now]][i];
q.push(nxt[now][i]);
}
}
}
}
int query(char buf[]) {
int len = strlen(buf);
int now = root;
int res = 0;
for(int i=0;i<len;i++) {
char c = buf[i] - 'a';
now = nxt[now][c];
int temp = now;
while(temp!=root) {
res += end[temp];
end[temp] = 0;
temp = fail[temp];
}
}
return res;
}
};
char buf[maxn*2];
Trie ac;
int main(){
int T;
scanf("%d",&T);
while(T--) {
int n;
scanf("%d", &n);
ac.init();
for(int i=0;i<n;i++) {
scanf("%s", buf);
ac.insert(buf);
}
ac.build();
scanf("%s",buf);
printf("%d\n", ac.query(buf));
}
return 0;
}
Bitset
#include<bits/stdc++.h>
using namespace std;
bitset<5> b, a;
int main() {
cout << b << endl; //00000
b.set(2);
cout << b << endl;//00100
a.set(3);
cout << a << endl;//01000
b ^= a;//(00100)^(01000)
cout << b << endl;//01100
b.reset(3);//00100
cout << b << endl;
b.flip();//11011
cout << b << endl;
}
最大子段和
int best=0,sum=0;
for(int i=0;i<n;i++)
{
sum=max(array[i],sum+array[k]);
best=max(sun,best);
}
cout<<best<<endl;
Codeforces977E dfs判环
187ms 用dfs找连通块,并且需要环上的每个点的度为2
#include<bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 5;
int vis[maxn];
int ans, f;
vector<int> e[maxn];
void dfs(int u) { //遍历每个联通块
vis[u] = 1;
if(e[u].size()!=2)
f = 1;
for(int v : e[u]) {
if(!vis[v])
dfs(v);
}
}
int main() {
int n, m;
scanf("%d%d",&n,&m);
for (int i = 1; i <= m;i++) {
int u,v;
scanf("%d%d",&u,&v);
e[v].push_back(u);
e[u].push_back(v);
}
for(int i=1;i<=n;i++) {
f = 0;
if(!vis[i]) {
dfs(i);
if(f==0)ans++;
}
}
printf("%d\n", ans);
}
93ms 并查集
#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 5;
int f[maxn];
int d[maxn];
struct edge{
int u, v;
}e[maxn];
int find(int x) {
return f[x] == x ? x : f[x] = find(f[x]);
}
int main() {
int n,m;scanf("%d%d",&n,&m);
for(int i=0;i<=n;i++)
f[i] = i;
for(int i=1;i<=m;i++) {
scanf("%d%d", &e[i].u, &e[i].v);
d[e[i].u]++;
d[e[i].v]++;
}
int ans = 0;
for(int i=1;i<=m;i++) {
if(d[e[i].u]==2&&d[e[i].v]==2) {
int a = find(e[i].u);
int b = find(e[i].v);
if(a==b)
ans++;
else if(a<b)
f[b] = a;
else
f[a] = b;
}
}
cout << ans << endl;
}
Codeforces[DP] k-Tree
寒假手速赛第一场
题意:...
时间复杂度:\(O(n*k)\)
思路:直接模拟，选或者不选，dp数组第一维为和，第二维为是否包含重量大于d的边
（考虑组合数做法？）
https://codeforces.com/group/GlRm4CeuZ9/contest/266331/problem/C
#include<bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const int maxn = 1e5 + 5;
long long dp[maxn][3];
int main() {
int n, k, d;
scanf("%d%d%d", &n, &k, &d);
dp[0][0] = 1;
for (int i = 1; i <= n;i++) {
for (int j = 1; j <= k;j++) {
if(i>=j) {
if(j<d) {
dp[i][0] = (dp[i][0] + dp[i - j][0]) % mod;
dp[i][1] = (dp[i][1] + dp[i - j][1]) % mod;
}
else if(j>=d) {
dp[i][1] = (dp[i][1] + dp[i - j][0] + dp[i - j][1]) % mod;
}
}
}
}
cout << dp[n][1] << endl;
}
Codeforces[DP] LCIS
寒假手速赛第一场
题意:求给定两个序列的公共最长子序列
时间复杂度:\(O(n^2)\)
思路:增加一个指针，一个记录路径的数组即可
https://codeforces.com/group/GlRm4CeuZ9/contest/266331/problem/D
#include<bits/stdc++.h>
using namespace std;
const int maxn = 500 + 5;
int a[maxn],s[maxn];
int dp[maxn], pre[maxn];
int main() {
int n;
scanf("%d", &n);
for (int i = 1; i <= n;i++){
scanf("%d", &a[i]);
}
int m;
scanf("%d", &m);
for (int i = 1; i <= m;i++) {
scanf("%d", &s[i]);
}
//
for (int i = 1; i <= n;i++) {
int x = 0;
for (int j = 1; j <= m; j++) {
if(a[i]<s[j])
continue;
if(a[i]>s[j]) {
if(dp[j]>dp[x]){
x = j;
}
}
if(a[i] == s[j]) {
if(dp[j]<dp[x]+1) {
dp[j] = dp[x] + 1;
pre[j] = x;
}
}
}
}
//
int ans = 0;
int t = 0;
for (int i = 1; i <= m;i++) {
if(dp[i]>dp[t]){
ans = dp[i];
t = i;
}
}
printf("%d\n", ans);
vector<int> v;
while(t) {
v.push_back(s[t]);
t = pre[t];
}
int sz = v.size();
for (int i = sz - 1; i >= 0;i--) {
printf("%d", v[i]);
if(i == 0) printf("\n");
else printf(" ");
}
}
Codeforces[DP] Elevator
https://codeforces.com/group/GlRm4CeuZ9/contest/266331/problem/E
题意：某电梯可以承载4个人，上、下电梯均需要1s，上、下一层均需要1s，乘客按照排队顺序上电梯
给定每个乘客所在楼层和目标楼层，问送完所有乘客的最短时间
#include<bits/stdc++.h>
using namespace std;
const int maxn = 2000 + 5;
const int inf = 0x3f3f3f3f;
int dp[maxn][10][10][10][10];
int x[maxn], y[maxn];
int n;
int dis(int x,int y) {
return abs(x - y);
}
int dfs(int i,int cur,int a,int b,int c) {
if(dp[i][cur][a][b][c])
return dp[i][cur][a][b][c];
int ans = inf;
if(i>n) {
if(!a&&!b&&!c)
return 0;
if(a)
ans = min(ans, dfs(i, a, 0, b, c) + dis(cur, a) + 1);
if(b)
ans = min(ans, dfs(i, b, a, 0, c) + dis(cur, b) + 1);
if(c)
ans = min(ans, dfs(i, c, a, b, 0) + dis(cur, c) + 1);
return dp[i][cur][a][b][c] = ans;
}
if(a)
ans = min(ans, dfs(i, a, 0, b, c) + dis(cur, a) + 1);
if(b)
ans = min(ans, dfs(i, b, a, 0, c) + dis(cur, b) + 1);
if(c)
ans = min(ans, dfs(i, c, a, b, 0) + dis(cur, c) + 1);
if(a&&b&&c) {
ans = min(ans, dfs(i + 1, y[i], a, b, c) + dis(cur, x[i]) + dis(x[i], y[i]) + 2);
ans = min(ans, dfs(i + 1, a, y[i], b, c) + dis(cur, x[i]) + dis(x[i], a) + 2);
ans = min(ans, dfs(i + 1, b, a, y[i], c) + dis(cur, x[i]) + dis(x[i], b) + 2);
ans = min(ans, dfs(i + 1, c, a, b, y[i]) + dis(cur, x[i]) + dis(x[i], c) + 2);
}
else {
if(!a)
ans = min(ans, dfs(i + 1, x[i], y[i], b, c) + dis(cur, x[i]) + 1);
else if(!b)
ans = min(ans, dfs(i + 1, x[i], a, y[i], c) + dis(cur, x[i]) + 1);
else if(!c)
ans = min(ans, dfs(i + 1, x[i], a, b, y[i]) + dis(cur, x[i]) + 1);
}
return dp[i][cur][a][b][c] = ans;
}
int main() {
scanf("%d", &n);
for (int i = 1;i<=n;i++)
scanf("%d%d", &x[i], &y[i]);
printf("%d\n", dfs(1, 1, 0, 0, 0));
}
Codeforces[DP] 状压 Marbles
https://codeforces.com/contest/1215/problem/E
#include<bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const int maxn = 4e5 + 100;
int a[maxn];
int tot;
long long pre[30][30], cnt[30];
long long dp[1 << 21];
map<int, int> mp;
int main()
{
int n;
scanf("%d", &n);
for (int i = 1; i <= n;i++) {
scanf("%d", &a[i]);
if(!mp[a[i]]){
mp[a[i]] = tot++;
}
}
for (int i = 1; i <= n;i++) {
int x = mp[a[i]];
for (int j = 0; j < tot;j++) {
pre[x][j] += cnt[j];
}
cnt[x]++;
}
memset(dp, 0x3f, sizeof(dp));
dp[0] = 0;
for (int i = 1; i < (1 << tot); i++) {
for (int j = 0; j < tot;j++) {
if(i&(1<<j)) {
long long temp = 0;
for (int k = 0; k < tot;k++) {
if(j!=k&&!(i&(1<<k))) {
temp += pre[j][k];
}
}
dp[i] = min(dp[i], dp[i ^ (1 << j)] + temp);
}
}
}
printf("%lld\n", dp[(1 << tot) - 1]);
}
