D. Monitor Educational Codeforces Round 28

http://codeforces.com/contest/846/problem/D

二分答案

适合于：

判断在t时候第一次成立

哪个状态是最小代价

 #include <cstdio>
 #include <cstdlib>
 #include <cmath>
 #include <cstring>
 #include <time.h>
 #include <string>
 #include <set>
 #include <map>
 #include <list>
 #include <stack>
 #include <queue>
 #include <vector>
 #include <bitset>
 #include <ext/rope>
 #include <algorithm>
 #include <iostream>
 using namespace std;
 #define ll long long
 #define minv 1e-6
 #define inf 1e9
 #define pi 3.1415926536
 #define E  2.7182818284
 const ll mod=1e9+;//
 const int maxn=5e2+;

 struct node
 {
     int x,y,t;
 }f[maxn*maxn];

 int K,s,t,a[maxn][maxn],tot[maxn][maxn]={};

 int cmp(node a,node b)
 {
     return a.t<b.t;
 }

 bool judge()
 {
     int i,j;
     for (i=K;i<=s;i++)
         for (j=K;j<=t;j++)
             if (tot[i][j]-tot[i-K][j]-tot[i][j-K]+tot[i-K][j-K]==K*K)
                 return ;
     return ;
 }

 int main()
 {
     int q,i,j,l,r,m;
     scanf("%d%d%d%d",&s,&t,&K,&q);
     for (i=;i<=q;i++)
         scanf("%d%d%d",&f[i].x,&f[i].y,&f[i].t);

     sort(f+,f+q+,cmp);

     l=,r=q;
     while (l<=r)
     {
         memset(a,,sizeof(a));
         m=(l+r)>>;

         for (i=;i<=m;i++)
             a[f[i].x][f[i].y]=;
         for (i=;i<=s;i++)
             for (j=;j<=t;j++)
                 tot[i][j]=tot[i-][j]+tot[i][j-]-tot[i-][j-]+a[i][j];
         if (judge())
             r=m-;
         else
             l=m+;
     }
     if (l==q+)
         printf("-1");
     else
         printf("%d",f[l].t);
     return ;
 }
 /*
 2 2 2 4
 1 2 1
 1 1 4
 2 2 3
 2 1 2

 2 2 1 4
 1 2 1
 1 1 4
 2 2 3
 2 1 2

 2 2 2 0
 */