2017-2018 ACM-ICPC Southwestern European Regional Programming Contest (SWERC 2017)
A. Cakey McCakeFace
按题意模拟即可。
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { }
#define MS(x, y) memset(x, y, sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
const int N = 0, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei;
int n, m;
int a[2005], b[2005];
int main()
{
while(~scanf("%d%d",&n, &m))
{
for(int i = 1; i <= n; ++i)
{
scanf("%d", &a[i]);
}
for(int i = 1; i <= m; ++i)
{
scanf("%d", &b[i]);
}
map<int,int>mop;
int cnt = -1, val;
for(int i = 1; i <= n; ++i)
{
for(int j = 1; j <= m; ++j)if(b[j] >= a[i])
{
++mop[b[j] - a[i]];
}
}
for(auto it : mop)
{
if(it.second > cnt)
{
cnt = it.second;
val = it.first;
}
}
printf("%d\n", val);
}
return 0;
}
/*
【trick&&吐槽】
【题意】
【分析】
【时间复杂度&&优化】
*/
B. Table
枚举下底边，求出每个位置向上延伸的最大长度，枚举每个位置作为右下角，那么单调栈中每一个子矩形都可以对长宽都不超过它的询问产生贡献，通过差分二维前缀和，那么$O(1)$单点修改即可。
为了避免枚举单调栈中每一项，可以在每一项退栈时计算有多少个右下角可以与它产生贡献，这是对差分后的二维前缀和的一段区间加，再次差分前缀和转化成单点修改即可。
时间复杂度$O(xy+n+d)$。
#include<cstdio>
const int N=2010;
int n,m,_a,_b,i,j,a[N][N],b[N][N],f[N],q[N],t;
inline void work(int x,int y,int z,int o){
/*for(int i=y;i<=z;i++){
b[i-y][o]--;
b[i+1-x][o]++;
}
return;*/
b[0][o]--;
b[z-y+1][o]++;
b[y+1-x][o]++;
b[z+1-x+1][o]--;
}
int main(){
scanf("%d%d%d%d",&n,&m,&_a,&_b);
while(_a--){
int xl,xr,yl,yr;
scanf("%d%d%d%d",&xl,&xr,&yl,&yr);
a[xr][yr]++;
a[xl][yr]--;
a[xr][yl]--;
a[xl][yl]++;
}
for(i=n;i;i--)for(j=m;j;j--)a[i][j]+=a[i+1][j]+a[i][j+1]-a[i+1][j+1];
for(i=1;i<=n;i++)for(j=1;j<=m;j++)a[i][j]=a[i][j]==0;
/*puts("");
for(i=1;i<=n;i++){
for(j=1;j<=m;j++)printf("%d",a[i][j]);
puts("");
}*/
for(i=1;i<=n;i++){
for(j=1;j<=m;j++)if(a[i][j])f[j]++;else f[j]=0;
for(q[t=0]=0,j=1;j<=m;q[++t]=j++){
while(t&&f[j]<f[q[t]]){
work(q[t-1]+1,q[t],j-1,f[q[t]]);
t--;
}
}
for(j=1;j<=t;j++)work(q[j-1]+1,q[j],m,f[q[j]]);
}
for(j=1;j<=n;j++)for(i=1;i<=m+5;i++)b[i][j]+=b[i-1][j];
for(i=m;i;i--)for(j=n;j;j--)b[i][j]+=b[i+1][j]+b[i][j+1]-b[i+1][j+1];
while(_b--)scanf("%d%d",&i,&j),printf("%d\n",b[j][i]);
}
/*
7 5 3 9
1 2 0 1
5 7 2 5
0 1 2 4
7 1
3 5
5 3
2 2
3 3
4 4
4 5
6 2
1 1
*/
C. Macarons
设$f[i][S]$表示考虑前$i$行，第$i$行每个位置是否被填充的情况为$S$的合法方案数。
逐格转移预处理出转移矩阵后快速幂即可。
时间复杂度$O(2^{3n}\log m)$。
#include<cstdio>
#define rep(i) for(int i=0;i<m;i++)
typedef long long ll;
const int N=260,P=1000000000;
int n,m,S,i,j,k,x,y,z,f[N],g[N];
ll K;
int A[N][N],B[N][N],C[N][N];
inline void up(int&a,int b){a=a+b<P?a+b:a+b-P;}
inline void mul(int A[][N],int B[][N],int F[][N]){
rep(i)rep(j)C[i][j]=0;
rep(i)rep(j)if(A[i][j])rep(k)if(B[j][k])C[i][k]=(1LL*A[i][j]*B[j][k]+C[i][k])%P;
rep(i)rep(j)F[i][j]=C[i][j];
}
int main(){
scanf("%d%lld",&n,&K);
m=1<<n;
for(S=0;S<m;S++){
for(j=0;j<m;j++)f[j]=g[j]=0;
f[S]=1;
for(i=0;i<n;i++){
for(j=0;j<m;j++)g[j]=0;
for(j=0;j<m;j++)if(f[j]){
if(!(j>>i&1)){
up(g[j^(1<<i)],f[j]);
}else{
up(g[j^(1<<i)],f[j]);
up(g[j],f[j]);
if(i&&!(j>>(i-1)&1))up(g[j|(1<<(i-1))],f[j]);
}
}
for(j=0;j<m;j++)f[j]=g[j];
}
for(j=0;j<m;j++)A[j][S]=f[j];
}
for(i=0;i<m;i++)B[i][i]=1;
for(;K;K>>=1,mul(A,A,A))if(K&1)mul(A,B,B);
printf("%d",B[m-1][m-1]);
}
D. Candy Chain
设$f[i][j]$表示将区间$[i,j]$全部消完的最大收益，$dp[i]$表示考虑前$i$个位置的最大收益，那么$dp[i]=\max(dp[i-1],dp[j-1]+f[j][i](1\leq j\leq i))$。
对于$f$的计算，设$g[i][j][k]$表示考虑区间$[i,j]$，最后一次要删去的串在字典树上位于$k$点的最大收益，要么枚举一个区间消掉，要么在字典树上往下走，要么消去$k$代表的字符串。
时间复杂度$O(n^4c)$，但常数很小，且无用状态很多，远远达不到上界。
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=55,M=20005;
int n,m,i,j,k,x,tot,son[M][26],w[M],f[N][N],dp[N];
int g[N][M];
char a[N],b[N];
inline void up(int&a,int b){a<b?(a=b):0;}
inline void ins(int l,int v){
int x=0;
for(int i=0;i<l;i++){
int o=b[i]-'a';
if(!son[x][o])son[x][o]=++tot;
x=son[x][o];
}
up(w[x],v);
}
int main(){
for(i=1;i<M;i++)w[i]=-1;
scanf("%s",a+1);
n=strlen(a+1);
for(i=1;i<=n;i++)a[i]-='a';
scanf("%d",&m);
while(m--){
int l,w;
scanf("%s%d",b,&w);
l=strlen(b);
ins(l,w);
reverse(b,b+l);
ins(l,w);
}
for(i=0;i<=n+1;i++)for(j=0;j<=n+1;j++)f[i][j]=-1;
for(i=n;i;i--){
for(j=i-1;j<=n+1;j++)for(k=0;k<=tot;k++)g[j][k]=-1;
g[i-1][0]=0;
for(j=i-1;j<=n;j++)for(k=0;k<=tot;k++)if(~g[j][k]){
for(x=j+1;x<=n;x++)if(~f[j+1][x])up(g[x][k],g[j][k]+f[j+1][x]);
if(j<n){
int y=son[k][a[j+1]];
if(y){
up(g[j+1][y],g[j][k]);
if(~w[y])up(g[j+1][0],g[j][k]+w[y]);
}
}
if(k==0)up(f[i][j],g[j][k]);
}
}
for(i=1;i<=n;i++){
dp[i]=dp[i-1];
for(j=1;j<=i;j++)if(~f[j][i])up(dp[i],dp[j-1]+f[j][i]);
}
printf("%d",dp[n]);
}
E. Ingredients
拓扑排序之后背包即可。
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { }
#define MS(x, y) memset(x, y, sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
const int N = 1e6 + 10, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei;
int B, n;
char s1[N][24];
char s2[N][24];
char s3[N][24];
int c[N];
int v[N];
int cc[N];
int vv[N];
int ind[N];
int sta[N];
map<string, int>mop;
int ID;
vector< pair<int,int> >a[N];
int getid(char s[24])
{
if(!mop.count(s))
{
mop[s] = ++ID;
a[ID].clear();
cc[ID] = inf;
vv[ID] = -inf;
ind[ID] = 0;
}
return mop[s];
}
bool e[N];
LL f[10005];
int main()
{
while(~scanf("%d%d", &B, &n))
{
ID = 0;
mop.clear();
for(int i = 1; i <= n; ++i)
{
scanf("%s%s%s%d%d", s1[i], s2[i], s3[i], &c[i], &v[i]);
int z = getid(s1[i]);
int x = getid(s2[i]);
int y = getid(s3[i]);
++ind[z];
a[x].push_back({y, i});
a[y].push_back({x, i});
e[i] = 0;
}
int top = 0;
for(int i = 1; i <= ID; ++i)if(ind[i] == 0)
{
sta[++top] = i;
cc[i] = vv[i] = 0;
}
while(top)
{
int x = sta[top--];
for(auto it : a[x])
{
int y = it.first;
int o = it.second;
if(!e[o] && ind[y] == 0)
{
e[o] = 1;
int z = mop[s1[o]];
int cost = cc[x] + cc[y] + c[o];
int val = vv[x] + vv[y] + v[o];
if(cost < cc[z] || cost == cc[z] && val > vv[z])
{
cc[z] = cost;
vv[z] = val;
}
if(--ind[z] == 0)
{
sta[++top] = z;
}
}
}
}
MS(f, -63); f[0] = 0;
for(int i = 1; i <= ID; ++i)
{
//
//printf("cc = %d, vv = %d\n", cc[i], vv[i]);
//
for(int j = B; j >= cc[i]; --j)
{
gmax(f[j], f[j - cc[i]] + vv[i]);
}
}
int cost = 0;
LL val = 0;
for(int i = 1; i <= B; ++i)
{
if(f[i] > val)
{
val = f[i];
cost = i;
}
}
printf("%lld\n%d\n", val, cost);
}
return 0;
}
/*
【trick&&吐槽】
6 2
pizza_tomato pizza_base tomato 1 2
pizza_classic pizza_tomato cheese 5 5
7 2
pizza_tomato pizza_base tomato 1 2
pizza_classic pizza_tomato cheese 5 5
9 2
pizza_tomato pizza_base tomato 1 2
pizza_classic pizza_tomato cheese 5 5
100 2
pizza_tomato pizza_base tomato 1 2
pizza_classic pizza_tomato cheese 5 5
【题意】
【分析】
【时间复杂度&&优化】
*/
F. Shattered Cake
$ans=\frac{\sum_{i=1}^n w_il_i}{W}$。
#include<cstdio>
int w,n;long long s;
int main(){
scanf("%d%d",&w,&n);
while(n--){
int x,y;
scanf("%d%d",&x,&y);
s+=x*y;
}
printf("%lld",s/w);
}
G. Cordon Bleu
考虑费用流建图，新建一个容量为$n-1$的虚点，表示从餐厅出发的快递员，剩下部分显然，然后求最小费用最大流。
这样边数为$O(n^2)$，不能接受。注意到这些边费用都是曼哈顿距离，将绝对值拆开后$4$个象限分别用可持久化线段树优化建图即可。
如此一来点数边数均为$O(n\log n)$，可以通过。
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { }
#define MS(x, y) memset(x, y, sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
const int N = 70000, M = 1e6 + 10, Z = 1e9 + 7, inf = 1e9;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei;
int id, ST, ED;
int first[N];
int w[M], c[M], cap[M], cost[M], nxt[M];
int f[N];
int pe[N];
bool e[N];
int tot;
int FLAG=0;
void ins(int x, int y, int cap_, int cost_)
{
if(!x||!y)return;
if(FLAG)swap(x,y);
id ++;
w[id] = y;
cap[id] = cap_;
cost[id] = cost_;
nxt[id] = first[x];
first[x] = id;
id ++;
w[id] = x;
cap[id] = 0;
cost[id] = -cost_;
nxt[id] = first[y];
first[y] = id;
}
int q[N];
unsigned short head,tail;
void inq(int x, int cost_)
{
if(cost_ >= f[x]) return;
f[x] = cost_;
//pe[x] = pe_;
if(e[x]) return; e[x] = 1;
// if(cost_<q[head])q[--head]=x;else
q[++tail]=x;
}
int vis[N], tim;
int dfs(int x, int all)
{
if(x == ST) return all;
int use = 0;
vis[x] = tim;
for(int z = first[x]; z; z = nxt[z]) if(cap[z ^ 1]){
int y = w[z];
if(vis[y] != tim && f[y] + cost[z ^ 1] == f[x]){
int tmp = dfs(y, min(cap[z ^ 1], all - use));
cap[z ^ 1] -= tmp;
cap[z] += tmp;
use += tmp;
if(use == all) break;
}
}
return use;
}
bool spfa()
{
for(int i=0;i<=tot+10;i++)f[i]=inf;
cap[0] = inf;
head=1,tail=0;
inq(ST, 0);
while(head!=tail+1){
int x = q[head++]; e[x]=0;
for(int z = first[x]; z; z = nxt[z]){
if(cap[z]) inq(w[z], f[x] + cost[z]);
}
}
return f[ED] < inf;
}
int MCMF()
{
int maxflow = 0;
int mincost = 0;
while(spfa()){
int flow;
while(++tim, flow = dfs(ED, inf))
{
maxflow += flow;
mincost += f[ED] * flow;
}
}
//printf("%d\n",maxflow);
return mincost;
}
/*int MCMF()
{
int maxflow = 0;
int mincost = 0;
while(spfa()){
int flow = ~0U>>1;
int x = ED;
while(x != ST){
gmin(flow, cap[pe[x]]);
x = w[pe[x] ^ 1];
}
maxflow += flow;
mincost += f[ED] * flow;
x = ED;
while(x != ST){
cap[pe[x]] -= flow;
cap[pe[x] ^ 1] += flow;
x = w[pe[x] ^ 1];
}
}
return mincost;
//printf("%d %d\n", maxflow, mincost);
}*/
struct point
{
int x, y;
};
int dis(point a, point b)
{
return abs(a.x - b.x) + abs(a.y - b.y);
}
int n, m;
point a, b[N], cc[N];
int ans;
int pool[N];
inline bool cmpx(int x,int y){
return b[x].x<b[y].x;
}
int root0[1111],root1[1111];
int sonl[200000],sonr[200000];
int Ins(int x,int a,int b,int c,int p,int w){
int y=++tot;
if(a==b){
ins(y,x,inf,0);
ins(y,p,inf,w);
return y;
}
int mid=(a+b)>>1;
if(c<=mid){
sonl[y]=Ins(sonl[x],a,mid,c,p,w);
sonr[y]=sonr[x];
}else{
sonl[y]=sonl[x];
sonr[y]=Ins(sonr[x],mid+1,b,c,p,w);
}
ins(y,sonl[y],inf,0);
ins(y,sonr[y],inf,0);
return y;
}
void ask(int x,int a,int b,int c,int d,int p,int w){
if(!x)return;
if(c<=a&&b<=d){
ins(p,x,inf,w);
return;
}
int mid=(a+b)>>1;
if(c<=mid)ask(sonl[x],a,mid,c,d,p,w);
if(d>mid)ask(sonr[x],mid+1,b,c,d,p,w);
}
void solve()
{
MS(first, 0); id = 1;
ED = m + n + 1;
ST = m + n + 4;
int R = m + n + 2;
for(int i = 1; i <= m; i ++)
{
ins(ST, i, 1, 0);
/*for(int j = 1; j <= n; j ++)
{
ins(i, j + m, 1, dis(cc[i], b[j]));
}*/
}
tot=ST;
for(int i=1;i<=n;i++)pool[i]=i;
sort(pool+1,pool+n+1,cmpx);
root0[0]=root1[0]=0;
for(int i=1;i<=n;i++){
int x=pool[i];
root0[i]=Ins(root0[i-1],0,2000,b[x].y,x+m,-b[x].x-b[x].y);
root1[i]=Ins(root1[i-1],0,2000,b[x].y,x+m,-b[x].x+b[x].y);
}
for(int i=1;i<=m;i++){
int j=0;
int X=cc[i].x,Y=cc[i].y;
while(j<n&&b[pool[j+1]].x<=X)j++;
ask(root0[j],0,2000,0,Y,i,X+Y);
ask(root1[j],0,2000,Y,2000,i,X-Y);
}
reverse(pool+1,pool+n+1);
for(int i=1;i<=n;i++){
int x=pool[i];
root0[i]=Ins(root0[i-1],0,2000,b[x].y,x+m,b[x].x-b[x].y);
root1[i]=Ins(root1[i-1],0,2000,b[x].y,x+m,b[x].x+b[x].y);
}
for(int i=1;i<=m;i++){
int j=0;
int X=cc[i].x,Y=cc[i].y;
while(j<n&&b[pool[j+1]].x>=X)j++;
ask(root0[j],0,2000,0,Y,i,-X+Y);
ask(root1[j],0,2000,Y,2000,i,-X-Y);
}
for(int i = 1; i <= n; i ++)
{
ins(R, m + i, 1, dis(a, b[i]));
ins(m + i, ED, 1, 0);
}
ins(ST, R, n - 1 , 0);
//printf("%d %d\n",tot,id);return;
if(FLAG)swap(ST,ED);
ans += MCMF();
printf("%d\n", ans);
}
inline void getnum(int &x){
scanf("%d",&x);
//x=rand()%2001-1000;
x+=1000;
}
int main()
{
scanf("%d%d", &n, &m);//huowu ren
for(int i = 1; i <= n; i ++){
getnum(b[i].x);
getnum(b[i].y);
}
for(int i = 1; i <= m; i ++){
getnum(cc[i].x);
getnum(cc[i].y);
}
getnum(a.x);
getnum(a.y);
ans = 0;
for(int i = 1; i <= n; ++i)
{
ans += dis(a, b[i]);
}
solve();
return 0;
}
/*
【trick&&吐槽】
3 10
0 0
10 0
0 10
【题意】
【分析】
【时间复杂度&&优化】
*/
H. Kabobs
留坑。
I. Burglary
设$f[i][j][k]$表示考虑从上往下前$i$行，目前两条路线分别位于第$i$行的第$j$和第$k$根管道上的最大收益，暴力转移即可。
时间复杂度$O(10^4n)$。
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { }
#define MS(x, y) memset(x, y, sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
const int N = 1005, M = 5005, Z = 1e9 + 7, inf = 0x3f3f3f3f;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei;
int n, m;
char s[M];
int gv[N];
int v[N][M];
int p[N][M];
int gt[N];
int t[N][M];
int sum[N][M];
int f[N][12][12];
int main()
{
while (~scanf("%d%d", &n, &m))
{
for (int i = 1; i <= n; ++i)
{
scanf("%s", s + 1);
gv[i] = 0;
p[i][0] = 0;
v[i][0] = 0;
for (int j = 1; j <= m; ++j)if (isdigit(s[j]))
{
v[i][++gv[i]] = s[j] - '0';
p[i][gv[i]] = j;
}
++gv[i];
p[i][gv[i]] = m + 1;
v[i][gv[i]] = 0;
for (int j = 1; j <= gv[i]; ++j)
{
sum[i][j] = sum[i][j - 1] + v[i][j];
}
//
scanf("%s", s + 1);
gt[i] = 0;
for (int j = 1; j <= m; ++j)if (s[j] == '|')
{
t[i][++gt[i]] = j;
}
}
MS(f, -63);
MS(f[1], 0);
for (int o = 1; o < n; ++o)
{
for (int i = 1; i <= gt[o]; ++i)
{
for (int j = i; j <= gt[o]; ++j)if (f[o][i][j] >= 0)
{
for (int ii = 1; ii <= gt[o + 1]; ++ii)
{
for (int jj = ii; jj <= gt[o + 1]; ++jj)
{
int l1 = min(t[o][i], t[o + 1][ii]);
int r1 = max(t[o][i], t[o + 1][ii]);
int l2 = min(t[o][j], t[o + 1][jj]);
int r2 = max(t[o][j], t[o + 1][jj]);
int ll1 = lower_bound(p[o + 1], p[o + 1] + gv[o + 1] + 1, l1) - p[o + 1];
int rr1 = upper_bound(p[o + 1], p[o + 1] + gv[o + 1] + 1, r1) - p[o + 1] - 1;
int ll2 = lower_bound(p[o + 1], p[o + 1] + gv[o + 1] + 1, l2) - p[o + 1];
int rr2 = upper_bound(p[o + 1], p[o + 1] + gv[o + 1] + 1, r2) - p[o + 1] - 1;
//
int ll = max(ll1, ll2);
int rr = min(rr1, rr2);
if (ll <= rr)
{
int inside = sum[o + 1][rr]; if (ll)inside -= sum[o + 1][ll - 1];
if (inside)continue;
}
l1 = upper_bound(p[o + 1], p[o + 1] + gv[o + 1] + 1, l1) - p[o + 1] - 1;
r1 = lower_bound(p[o + 1], p[o + 1] + gv[o + 1] + 1, r1) - p[o + 1];
l2 = upper_bound(p[o + 1], p[o + 1] + gv[o + 1] + 1, l2) - p[o + 1] - 1;
r2 = lower_bound(p[o + 1], p[o + 1] + gv[o + 1] + 1, r2) - p[o + 1];
//
if (l1 > l2)
{
swap(l1, l2);
swap(r1, r2);
}
//one segment
int val;
if (l2 <= r1 + 1)
{
int l = l1;
int r = max(r1, r2);
val = sum[o + 1][r]; if (l)val -= sum[o + 1][l - 1];
}
//two segment
else
{
val = sum[o + 1][r1]; if (l1)val -= sum[o + 1][l1 - 1];
val += sum[o + 1][r2]; if (l2)val -= sum[o + 1][l2 - 1];
}
gmax(f[o + 1][ii][jj], f[o][i][j] + val);
}
}
}
}
}
int ans = 0;
for (int o = 1; o <= n; ++o)
{
for (int i = 1; i <= gt[o]; ++i)
{
for (int j = i; j <= gt[o]; ++j)if (f[o][i][j] >= 0)
{
int l = t[o][i];
int r = t[o][j];
l = upper_bound(p[o + 1], p[o + 1] + gv[o + 1] + 1, l) - p[o + 1] - 1;
r = lower_bound(p[o + 1], p[o + 1] + gv[o + 1] + 1, r) - p[o + 1];
int val = sum[o + 1][r]; if (l)val -= sum[o + 1][l - 1];
gmax(ans, f[o][i][j] + val);
}
}
}
printf("%d\n", ans);
}
return 0;
}
/*
【trick&&吐槽】
【题意】
【分析】
【时间复杂度&&优化】
5 20
--------------------
.|.....|...|......|.
1-1--1115-3-1-1--1-1
....|.........|.....
---1-11-1-11---1--3-
.......|.........|..
-7----9-4-----------
..|.................
--------9-----------
.|.................|
*/
J. Frosting on the Cake
按题意模拟即可。
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { }
#define MS(x, y) memset(x, y, sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
const int N = 0, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei;
int n;
LL a[3], b[3], c[3];
int main()
{
while(~scanf("%d",&n))
{
MS(a, 0); MS(b, 0); MS(c, 0);
for(int i = 1; i <= n; ++i)
{
int x; scanf("%d", &x);
a[i % 3] += x;
}
for(int i = 1; i <= n; ++i)
{
int x; scanf("%d", &x);
b[i % 3] += x;
}
for(int i = 0; i < 3; ++i)
{
for(int j = 0; j < 3; ++j)
{
c[(i + j) % 3] += a[i] * b[j];
}
}
printf("%lld %lld %lld\n", c[0], c[1], c[2]);
}
return 0;
}
/*
【trick&&吐槽】
【题意】
【分析】
【时间复杂度&&优化】
*/
K. Blowing Candles
求凸包后旋转卡壳，注意特判所有点共线的情况。
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { }
#define MS(x, y) memset(x, y, sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
const int N = 2e5 + 10, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei;
LL sqr(LL x)
{
return x * x;
}
struct point
{
LL x, y;
point(){}
point(LL x, LL y) : x(x), y(y) {}
friend point operator + (const point &a, const point &b){
return point(a.x + b.x, a.y + b.y);
}
friend point operator - (const point &a, const point &b){
return point(a.x - b.x, a.y - b.y);
}
friend point operator * (const point &a, const double &b){
return point(a.x * b, a.y * b);
}
friend point operator / (const point &a, const double &b){
return point(a.x / b, a.y / b);
}
friend bool operator == (const point &a, const point &b){
return a.x == b.x && a.y == b.y;
}
};
LL det(point a, point b)
{
return a.x * b.y - a.y * b.x;
}
LL dot(point a, point b)
{
return a.x * b.x + a.y * b.y;
}
LL dist(point a, point b)
{
return sqr(a.x - b.x) + sqr(a.y - b.y);
}
struct polygon_convex
{
vector<point> p;
polygon_convex(int size = 0){
p.resize(size);
}
};
bool comp_less(const point &a, const point &b)
{
return a.x - b.x < 0 || a.x - b.x == 0 && a.y - b.y < 0;
}
polygon_convex convex_hull(vector<point> a)
{
polygon_convex res(2 * a.size() + 5);
sort(a.begin(), a.end(), comp_less);
a.erase(unique(a.begin(), a.end()), a.end());
int m = 0;
for(int i = 0; i < a.size(); i ++){
while(m > 1 && det(res.p[m - 1] - res.p[m - 2], a[i] - res.p[m - 2]) <= 0) -- m;
res.p[m ++] = a[i];
}
int k = m;
for(int i = int(a.size()) - 2; i >= 0; i --){
while(m > k && det(res.p[m - 1] - res.p[m - 2], a[i] - res.p[m - 2]) <= 0) -- m;
res.p[m ++] = a[i];
}
res.p.resize(m);
if(a.size() > 1) res.p.resize(m - 1);
return res;
}
LL cross(point a, point b, point c)
{
return (c.x - a.x) * (b.y - a.y) - (b.x - a.x) * (c.y - a.y);
}
int R;
double rotating_calipers(point *p, int n)
{
int U = 1;
double ans = 2.0 * R;
p[n] = p[0];
for(int i = 0; i < n; i ++){
while(cross(p[i], p[i + 1], p[U + 1]) - cross(p[i], p[i + 1], p[U]) <= 0) U = (U + 1) % n;
double d = sqrt(dist(p[i], p[i + 1]));
double h = 1.0 * fabs(cross(p[i], p[i + 1], p[U])) / d;
gmin(ans, h);
}
return ans;
}
int n;
point t, p[N];
polygon_convex a;
int main()
{
scanf("%d%d", &n, &R);
for(int i = 0; i < n; i ++){
scanf("%lld%lld", &t.x, &t.y);
a.p.push_back(t);
}
a = convex_hull(a.p);
n = a.p.size();
for(int i = 0; i < n; i ++){
p[i] = a.p[i];
}
double ans;
if(n <= 2){
printf("0.000000000");
}
else {
ans = rotating_calipers(p, n);
printf("%.10f\n", ans);
}
return 0;
}
/*
【trick&&吐槽】
3 10
0 0
10 0
0 10
【题意】
【分析】
【时间复杂度&&优化】
*/
