Recently Luba bought a monitor. Mon

itor is a rectangular matrix of size?n?×?m. But then she started to notice that some pixels cease to work properly. Luba thinks that the monitor will become broken the first moment when it contains a square?k?×?kconsisting entirely of broken pixels. She knows that?q?pixels are already broken, and for each of them she knows the moment when it stopped working. Help Luba to determine when the monitor became broken (or tell that it's still not broken even after all?q?pixels stopped working).

Input

The first line contains four integer numbers?n,?m,?k,?q?(1?≤?n,?m?≤?500,?1?≤?k?≤?min(n,?m),?0?≤?q?≤?n·m)?— the length and width of the monitor, the size of a rectangle such that the monitor is broken if there is a broken rectangle with this size, and the number of broken pixels.

Each of next?q?lines contain three integer numbers?xi,?yi,?ti?(1?≤?xi?≤?n,?1?≤?yi?≤?m,?0?≤?t?≤?109)?— coordinates of?i-th broken pixel (its row and column in matrix) and the moment it stopped working. Each pixel is listed at most once.

We consider that pixel is already broken at moment?ti.

Output

Print one number — the minimum moment the monitor became broken, or "-1" if it's still not broken after these?q?pixels stopped working.

意思是求m*n中某k*k塊的顯示屏全部都壞了的話該顯示屏就壞了，求屏幕壞的t。

求k*k壞了可以用二位前綴和類似求面積的方法來

二維前綴和?https://blog.csdn.net/yzyyylx/article/details/78298318

因為時間是一直增加，可以二分時間求最小時間

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn=505;
struct node
{int x,y,t;
}s[500005];
int a[maxn][maxn],b[maxn][maxn];
int n,m,k,q;
bool cmp(node a,node b)
{return a.t<b.t;
}
int check(int t)//這里就表示二分的時間t時刻
{memset(a,0,sizeof(a));memset(b,0,sizeof(b));for(int i=1;i<=t;i++)a[s[i].x][s[i].y]=1;for(int i=1;i<=n;i++)//二位前綴和處理for(int j=1;j<=m;j++){b[i][j]=b[i-1][j]+b[i][j-1]-b[i-1][j-1]+a[i][j];}for(int i=k;i<=n;i++)//遍歷看是否成立for(int j=k;j<=m;j++){if(b[i][j]+b[i-k][j-k]-b[i-k][j]-b[i][j-k]==k*k)return 1;}return 0;
}
int main()
{int i,j;while(cin>>n>>m>>k>>q){for(i=1;i<=q;i++)cin>>s[i].x>>s[i].y>>s[i].t;sort(s+1,s+1+q,cmp);int l=1,r=q;int ans=-1;while(l<=r){int mid=(l+r)/2;if(check(mid)){ans=s[mid].t;r=mid-1;}elsel=mid+1;}printf("%d\n",ans);}return 0;
}

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