2020 ICPC Universidad Nacional de Colombia Programming Contest
A. Approach
三分
显然答案可以三分，注意\(eps\)还有两条线平行的情况
view code
#include<bits/stdc++.h>
using namespace std;
#pragma GCC optimize("O2")
double dis2(double ox, double oy, double ex, double ey){ return (ox - ex) * (ox - ex) + (oy - ey) * (oy - ey); }
const double eps = 1e-12;
typedef long long ll;
ll asx, asy, aex, aey;
ll bsx, bsy, bex, bey;
double alen, blen;
double check(double t){
double x1, y1, x2, y2, dx, dy, d;
dx = 1.0*aex - asx, dy = 1.0*aey - asy;
d = alen;
if(t>=d-eps) x1 = aex, y1 = aey;
else x1 = asx + t * (dx / d), y1 = asy + t * (dy / d);
dx = bex - bsx, dy = bey - bsy;
d = blen;
if(t>=d-eps) x2 = bex, y2 = bey;
else x2 = bsx + t * (dx / d), y2 = bsy + t * (dy / d);
return dis2(x1,y1,x2,y2);
}
int main(){
cin >> asx >> asy >> aex >> aey;
cin >> bsx >> bsy >> bex >> bey;
alen=sqrt(dis2(asx,asy,aex,aey)),blen=sqrt(dis2(bsx,bsy,bex,bey));
double l = 0, r = min(alen,blen);
double ans=1e25;
for(int i = 1; i <= 100000; i++){
double ml = (2*l+r) / 3;
double mr = (l+2*r) / 3;
double sl=check(ml),sr=check(mr);
if(sl+eps <= sr) r = mr;
else l = ml;
ans=min({ans,sl,sr});
}
l=min(alen,blen);r=max(alen,blen);
for(int i = 1; i <= 100000; i++){
double ml = (2*l+r) / 3;
double mr = (l+2*r) / 3;
double sl=check(ml),sr=check(mr);
if(sl+eps <= sr) r = mr;
else l = ml;
ans=min({ans,sl,sr});
}
cout << fixed << setprecision(12) << sqrt(ans) << endl;
return 0;
}
B. Baby name
\(SA\)
第二个串必然是选字典序最大的后缀，的一个串选择字典序最大的后缀的一个前缀，除了第一个字符之外所有字符都要比第二个串的首字母大或等
直接\(SA\)排序即可
view code
#include <bits/stdc++.h>
using namespace std;
#define LL long long
const int N=2e6+5,MOD=1e9+7;
int rk[N],sec[N],sa[N],c[N];
void SA(char *s,int n,int m){
for(int i=1;i<=m;i++)c[i]=0;
for(int i=1;i<=n;i++)c[rk[i]=(int)(s[i])]++;
for(int i=1;i<=m;i++)c[i]+=c[i-1];
for(int i=1;i<=n;i++)sa[c[rk[i]]--]=i;
for(int k=1;k<=n;k<<=1){
int p=0;
for(int i=n-k+1;i<=n;i++)sec[++p]=i;
for(int i=1;i<=n;i++)if(sa[i]>k)sec[++p]=sa[i]-k;
for(int i=1;i<=m;i++)c[i]=0;
for(int i=1;i<=n;i++)c[rk[sec[i]]]++;
for(int i=1;i<=m;i++)c[i]+=c[i-1];
for(int i=n;i>=1;i--)sa[c[rk[sec[i]]]--]=sec[i];
swap(rk,sec);
p=1;
rk[sa[1]]=1;
for(int i=2;i<=n;i++)rk[sa[i]]=(sec[sa[i]]==sec[sa[i-1]]&&sec[sa[i]+k]==sec[sa[i-1]+k]?p:++p);
if(p==m)break;
m=p;
}
}
char s[N],t[N];
int main()
{
// freopen("in.txt","r",stdin);
ios::sync_with_stdio(0);
cin>>s+1>>t+1;
int ls=strlen(s+1),lt=strlen(t+1);
SA(t,lt,300);
int pt=sa[lt];
SA(s,ls,300);
int ps=sa[ls];
string ans;
for(int i=ps;i<=ls;i++)
if(i==ps||s[i]>=t[pt])ans+=s[i];else break;
for(int i=pt;i<=lt;i++)ans+=t[i];
cout<<ans<<endl;
}
C. Cipher count
埃氏筛
一个串为需要被计数的串，当且仅当它的最小循环节为其本身
枚举每个长度，\(f(i)\)表示长度为\(i\)的且最小循环节为为自身的串的数量，那么可以得到\(f(i)=a^i - \sum_{d\mid i}f(d)\)
直接类似埃氏筛一样处理即可
view code
//#pragma GCC optimize("O3")
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
const int MOD = (int)1e9+7;
const int MAXN = 5e6+7;
int f[MAXN], a, k, ret;
void solve(){
scanf("%d %d",&a,&k);
f[0] = 1;
for(int i = 1; i <= k; i++) f[i] = 1ll * f[i-1] * a % MOD;
for(int i = 1; i <= k; i++) {
ret = (ret + f[i]) % MOD;
for(int j = i + i; j <= k; j += i) f[j] = (f[j] - f[i] + MOD) % MOD;
}
cout << ret << endl;
}
int main(){
#ifndef ONLINE_JUDGE
freopen("Local.in","r",stdin);
// freopen("ans.out","w",stdout);
#endif
solve();
return 0;
}
D. Dice
矩阵快速幂
view code
#include <bits/stdc++.h>
using namespace std;
#define LL long long
const int N=2e6+5,MOD=1e9+7;
int SIZ;
struct node{
int a[205][205];
node(){memset(a,0,sizeof a);}
int* operator[](int x){ return a[x]; }
friend node operator*(node a,node b){
node c;
for(int i=0;i<SIZ;i++){
for(int j=0;j<SIZ;j++){
for(int k=0;k<SIZ;k++){
c[i][j]=(c[i][j]+(LL)a[i][k]*b[k][j])%MOD;
}
}
}
return c;
}
}f,f0;
node qpow(node a,LL b){
node res;
for(int i=0;i<SIZ;i++)res[i][i]=1;
for(;b;b>>=1,a=a*a)
if(b&1)res=res*a;
return res;
}
LL qpow(LL a,LL b){
LL res=1;
for(;b;b>>=1,a=a*a%MOD)
if(b&1)res=res*a%MOD;
return res;
}
int w[N];
int main()
{
// freopen("in.txt","r",stdin);
ios::sync_with_stdio(0);
int n,k,m;
cin>>n>>k>>m;
int inv=qpow(k-k/m,MOD-2);
for(int i=1;i<=k;i++){
if(i%m==0)continue;
w[i%m]++;
}
for(int i=0;i<m;i++)w[i]=(LL)w[i]*inv%MOD;
SIZ=m;
f0[0][0]=1;
for(int i=0;i<m;i++){
for(int j=0;j<m;j++){
f[i][j]=w[(j-i+m)%m];
}
}
auto fto=qpow(f,n);
auto ans=f0*fto;
cout<<ans[0][0]<<endl;
}
E. Enter to the best problem of this contest!
交互
先询问\(1\)号点，记录距离，然后每次询问当前点的一个儿子，如果距离当前点的距离大就进入另一个儿子的子树，否则进入当前儿子的子树
view code
#include<bits/stdc++.h>
using namespace std;
int query(int x){
cout << x << endl;
cin >> x;
return x;
}
void print(int x){ cout << "! " << x <<endl; }
int main(){
int n;
scanf("%d",&n);
int u = 1;
int dis = query(1);
if(!dis){
print(1);
return 0;
}
for(int i = 2; i <= n; i++){
int ndis = query(u<<1);
if(dis==1){
if(ndis==2) print(u<<1|1);
else print(u<<1);
return 0;
}
if(ndis>dis) u = u << 1 | 1, dis = ndis - 2;
else u = u << 1, dis = ndis;
}
return 0;
}
F. Free restricted flights
分层图最短路
建正图和反图，对于两个点，分别正图反图跑一次最短路，\(dis[ta][tb][i][kk]\)表示人为\(a\)或\(b\)，跑的是正图或反图，起始点到\(i\)点，用了\(kk\)次\(ticket\)的最小花费
然后枚举每个点计算最小值即可
view code
#include<bits/stdc++.h>
using namespace std;
#define pii pair<int,int>
#define INF 0x3f3f3f3f
const int MAXN = 1e4+7;
int n, m, a, b, k;
int dis[2][2][MAXN][11];
vector<pii> G[2][MAXN];
void dijkstra(int ta, int tb, int s){
priority_queue<pair<int,pii>,vector<pair<int,pii> >, greater<pair<int,pii> > > que;
dis[ta][tb][s][0] = 0;
que.push(make_pair(0,make_pair(s,0)));
while(!que.empty()){
auto p = que.top(); que.pop();
int d = p.first, u = p.second.first, kk = p.second.second;
if(dis[ta][tb][u][kk]!=d) continue;
for(auto &e : G[tb][u]){
int v = e.first, w = e.second;
if(dis[ta][tb][v][kk] > dis[ta][tb][u][kk] + w){
dis[ta][tb][v][kk] = dis[ta][tb][u][kk] + w;
que.push(make_pair(dis[ta][tb][v][kk],make_pair(v,kk)));
}
if(kk!=k){
if(dis[ta][tb][v][kk+1] > dis[ta][tb][u][kk]){
dis[ta][tb][v][kk+1] = dis[ta][tb][u][kk];
que.push(make_pair(dis[ta][tb][v][kk+1],make_pair(v,kk+1)));
}
}
}
}
}
int main(){
scanf("%d %d %d %d %d",&n,&m,&a,&b,&k);
for(int i = 1; i <= m; i++){
int u, v, w;
scanf("%d %d %d",&u,&v,&w);
G[0][u].push_back(make_pair(v,w));
G[1][v].push_back(make_pair(u,w));
}
memset(dis,0x3f,sizeof(dis));
dijkstra(0,0,a);
dijkstra(0,1,a);
dijkstra(1,0,b);
dijkstra(1,1,b);
int u = INT_MAX;
int ret = INF;
for(int i = 0; i < n; i++) if(i!=a and i!=b){
int da = INF, db = INF;
for(int j = 0; j <= k; j++) for(int jj = 0; jj + j <= k; jj++){
if(dis[0][0][i][j]+dis[0][1][i][jj]<da) da = dis[0][0][i][j]+dis[0][1][i][jj];
if(dis[1][0][i][j]+dis[1][1][i][jj]<db) db = dis[1][0][i][j]+dis[1][1][i][jj];
}
if(da+db<ret){
ret = da + db;
u = i;
}
}
if(ret>=INF) cout << ">:(" << endl;
else cout << u << ' ' << ret << endl;
return 0;
}
G. Great dinner
组合数学
\(m\)对学生和剩下的\(n-2m\)个分别算
假设现在已经有\(x\)个学生在队列中，那么放入一对学生，枚举第一个学生在的位置，一共有\(\sum_{i=1}^{x+1}i\)种安置方法法
放入一个学生，则有\(x+1\)种安置方法，乘起来即可
view code
#include<bits/stdc++.h>
using namespace std;
int n, m;
const int MOD = 1e9+7;
long long cal(int x){ return (1ll + x) * x / 2 % MOD; }
int main(){
scanf("%d %d",&n,&m);
for(int i = 1; i <= m; i++){
int x, y;
scanf("%d %d",&x,&y);
}
long long ret = 1;
int tot = 0;
for(int i = 1; i <= m; i++){
ret = ret * cal(tot+1) % MOD;
tot += 2;
}
int num = n - 2 * m;
for(int i = 1; i <= num; i++) ret = ret * (tot + 1ll) % MOD, tot += 1;
cout << ret << endl;
return 0;
}
H. Happy game
\(PAM\)板子题
答案就是本质不同的长度大于\(1\)的奇数回文子串的数量
view code
#include <bits/stdc++.h>
using namespace std;
#define LL long long
const int N=6e5+5,MOD=1e9+7;
class PAM{
public:
int tot,sz[N],len[N],ch[N][26],fail[N],n,last,ans;
char s[N];
void init(){
cin>>n>>s+1;
len[0]=0;
len[1]=-1;
fail[0]=1;
fail[1]=0;
tot=1;
last=0;
ans=0;
}
int getfail(int x,int pos){
while(s[pos]!=s[pos-len[x]-1])x=fail[x];
return x;
}
void insert(int x,int pos){
int u=getfail(x,pos);
int c=s[pos]-'a';
if(!ch[u][c]){
len[++tot]=len[u]+2;
fail[tot]=ch[getfail(fail[u],pos)][c];
sz[tot]=sz[fail[tot]]+1;
ch[u][c]=tot;
if(len[tot]>=3&&(len[tot]&1))ans++;
}
last=ch[u][c];
}
void build(){
for(int i=1;i<=n;i++){
insert(last,i);
}
}
}t;
int main()
{
// freopen("H.in","r",stdin);
ios::sync_with_stdio(0);
t.init();
t.build();
cout<<t.ans<<endl;
return 0;
}
I. Incredible photography
单调栈
显然我们需要从高度最高的位置开始枚举来计算答案，对于枚举的每个位置，分别找到其左边和右边比当前高度高的第一个建筑，如果存在高度相同的要继续往两边找（为了保证走的最远），记位置\(i\)的答案是\(f(i)\)，找到的左边位置为\(l\)，右边位置为\(r\)，则\(f(i) = \max(f(l)+dis(l,i),f(r)+dis(i,r))\)
view code
#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1e5+7;
#define LL long long
int n, A[MAXN], last[MAXN];
int B[MAXN][20], lp[MAXN], rp[MAXN], lt[MAXN], rt[MAXN];
LL f[MAXN];
int stk[MAXN], top;
vector<int> vec;
int main(){
scanf("%d",&n);
for(int i = 1; i <= n; i++) scanf("%d",&A[i]), vec.push_back(A[i]);
sort(vec.begin(),vec.end()); vec.erase(unique(vec.begin(),vec.end()),vec.end());
for(int i = 1; i <= n; i++) A[i] = lower_bound(vec.begin(),vec.end(),A[i]) - vec.begin() + 1;
for(int i = 1; i <= n; i++) B[i][0] = A[i];
for(int j = 1; (1 << j) <= n; j++) for(int i = 1; i + (1 << j) - 1 <= n; i++){
B[i][j] = max(B[i][j-1],B[i+(1<<(j-1))][j-1]);
}
auto qmax = [&](int L, int R){
if(L>R) return -1;
int d = log2(R - L + 1);
return max(B[L][d],B[R-(1<<d)+1][d]);
};
memset(last,0,sizeof(last));
for(int i = 1; i <= n; i++){
if(!last[A[i]] or qmax(last[A[i]],i)>A[i]) lp[i] = i;
else lp[i] = lp[last[A[i]]];
last[A[i]] = i;
}
memset(last,0,sizeof(last));
for(int i = n; i >= 1; i--){
if(!last[A[i]] or qmax(i,last[A[i]])>A[i]) rp[i] = i;
else rp[i] = rp[last[A[i]]];
last[A[i]] = i;
}
top = 0;
for(int i = 1; i <= n; i++){
lt[i] = i;
while(top and A[i]>=A[stk[top]]) lt[i] = lt[stk[top--]];
stk[++top] = i;
}
top = 0;
for(int i = n; i >= 1; i--){
rt[i] = i;
while(top and A[i]>=A[stk[top]]) rt[i] = rt[stk[top--]];
stk[++top] = i;
}
// for(int i = 1; i <= n; i++) cout << lt[i] << ' ' <<rt[i] <<endl;
vector<pair<int,int> > vcc;
for(int i = 1; i <= n; i++) vcc.push_back(make_pair(A[i],i));
sort(vcc.begin(),vcc.end(),greater<pair<int,int> >());
for(int i = 0; i < n; i++){
int id = vcc[i].second;
// cout << id << ' ' << lt[i] - 1 << ' ' << rt[i] + 1 <<endl;
if(lt[id]-1!=0){
int p = lt[id] - 1;
f[id] = max(f[id],f[lp[p]] + id - lp[p]);
}
if(rt[id]+1!=n+1){
int p = rt[id] + 1;
f[id] = max(f[id],f[rp[p]] + rp[p] - id);
}
}
for(int i = 1; i <= n; i++) cout << f[i] << " \n"[i==n];
return 0;
}
J. Java exam
K. Katastrophic sort
view code
#include <bits/stdc++.h>
using namespace std;
#define LL long long
const int N=2e6+5,MOD=1e9+7;
char s[N],t[N],s1[N],t1[N];
int main()
{
// freopen("in.txt","r",stdin);
ios::sync_with_stdio(0);
cin>>s>>t;
int n=strlen(s),l2=strlen(t);
if(strcmp(s,t)==0){
cout<<"="<<endl;
return 0;
}
int p1=0,p2=0;
for(int i=0;i<n;i++){
if(p1<n){
if(s[p1]>='a'&&s[p1]<='z'){
if(s[p1]==t[p2]){
p1++,p2++;
continue;
}
else {
if(s[p1]<t[p2]){
cout<<"<"<<endl;
return 0;
}
else {
cout<<">"<<endl;
return 0;
}
}
}
else{
int t1=p1,t2=p2;
while(t1<n&&s[t1]>='0'&&s[t1]<='9')t1++;
while(t2<l2&&t[t2]>='0'&&t[t2]<='9')t2++;
if(t1-p1!=t2-p2){
if(t1-p1<t2-p2){
cout<<"<"<<endl;
return 0;
}
else {
cout<<">"<<endl;
return 0;
}
}
else{
for(int i=0;i<t1-p1;i++){
if(s[p1+i]<t[p2+i]){
cout<<"<"<<endl;
return 0;
}
else if(s[p1+i]>t[p2+i]){
cout<<">"<<endl;
return 0;
}
}
p1=t1,p2=t2;
}
}
}
else break;
}
}
L. Lonely day
\(BFS\)
预处理出每个方向连续块的数量
对于每个点向四周连边然后跑\(BFS\)同时记录边的方向即可
view code
#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for(int i=(a);i<=(n);i++)
#define per(i,a,n) for(int i=(n);i>=(a);i--)
#define LL long long
const int N=6e6+5,MOD=1e9+7;
char s[2005][2005];
int to[4][2005][2005],dis[2005][2005],pre[2005][2005],px[2005][2005],py[2005][2005];
struct node{
int x,y;
}src,des,que[N];
int st,ed;
int dx[]={1,0,0,-1};
int dy[]={0,-1,1,0};
string ans;
void dfs(int x,int y){
if(src.x==x && src.y==y){
return;
}
if(pre[x][y]==0)ans+='D';
else if(pre[x][y]==1)ans+='L';
else if(pre[x][y]==2)ans+='R';
else if(pre[x][y]==3)ans+='U';
dfs(px[x][y],py[x][y]);
}
int main()
{
// freopen("in.txt","r",stdin);
ios::sync_with_stdio(0);
int n,m;
cin>>n>>m;
rep(i,1,n){
cin>>s[i]+1;
rep(j,1,m){
if(s[i][j]=='S')src=(node){i,j};
else if(s[i][j]=='E')des=(node){i,j},s[i][j]='.';
dis[i][j]=-1;
}
}
rep(i,1,n){
int las=0;
rep(j,1,m){
to[1][i][j]=las+1;
if(s[i][j]=='X')las++;else las=0;
}
las=0;
per(j,1,m){
to[2][i][j]=las+1;
if(s[i][j]=='X')las++;else las=0;
}
}
rep(j,1,m){
int las=0;
rep(i,1,n){
to[3][i][j]=las+1;
if(s[i][j]=='X')las++;else las=0;
}
las=0;
per(i,1,n){
to[0][i][j]=las+1;
if(s[i][j]=='X')las++;else las=0;
}
}
st=1,ed=0;
que[++ed]=src;dis[src.x][src.y]=0;
while(st<=ed){
auto u=que[st++];
rep(i,0,3){
int nx=to[i][u.x][u.y]*dx[i]+u.x,ny=to[i][u.x][u.y]*dy[i]+u.y;
if(nx>=1&&ny>=1&&nx<=n&&ny<=m&&s[nx][ny]=='.'&&dis[nx][ny]==-1){
dis[nx][ny]=dis[u.x][u.y]+1;
pre[nx][ny]=i;
px[nx][ny]=u.x;
py[nx][ny]=u.y;
que[++ed]=(node){nx,ny};
}
}
}
int res=dis[des.x][des.y];
cout<<res<<endl;
if(~res){
dfs(des.x,des.y);
reverse(ans.begin(),ans.end());
cout<<ans<<endl;
}
return 0;
}
M. Magic spells
子序列自动机
用原串建子序列自动机，然后每个询问串在上面跑即可
view code
#include <bits/stdc++.h>
using namespace std;
#define LL long long
const int N=2e6+5,MOD=1e9+7;
char s[N],t[N];
int las[N],nxt[(int)2e5+5][27];
int main()
{
// freopen("in.txt","r",stdin);
ios::sync_with_stdio(0);
cin>>s+1;
int n=strlen(s+1);
for(int i=n;i>=1;i--){
for(int j=0;j<26;j++)nxt[i][j]=las[j];
las[s[i]-'a']=i;
}
for(int j=0;j<26;j++){
nxt[0][j]=las[j];
}
int m;
cin>>m;
for(int i=1;i<=m;i++){
cin>>t;
int l2=strlen(t);
int now=0,cnt=0,ok=1;
for(int j=0;ok&&j<l2;j++){
if(nxt[now][t[j]-'a']>now){
cnt++;
now=nxt[now][t[j]-'a'];
}
else ok=0;
}
if(!cnt){
cout<<"IMPOSSIBLE\n";
}
else{
for(int j=0;j<cnt;j++){
cout<<t[j];
}
cout<<'\n';
}
}
}
