Codeforces 846D Monitor（二維前綴和）

D. Monitor
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Recently Luba bought a monitor. Monitor 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 × k consisting 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.

Examples
input
2 3 2 5
2 1 8
2 2 8
1 2 1
1 3 4
2 3 2
output
8
input
3 3 2 5
1 2 2
2 2 1
2 3 5
3 2 10
2 1 100
output
-1

題目大意：

　　有一個N×M 的矩陣，給你Q 個位置壞的時間，如果有一個K×K 區域全部損壞則稱整個矩陣壞了。問矩陣由好變壞的時刻。

解題思路：

　　首先我們可以二分答案，對於一次cheak，由於我們要找檢查K×K 的子矩陣，我們可以先把時間小於等於二分值的位置制1 ，然後求二維前綴和，枚舉右下角用二維前綴和檢查即可。
　　總時間複雜度O(N×M×logMAXT)

AC代碼：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <stack>
using namespace std;
#define INF 0x3f3f3f3f3f3f3f3f
#define LL long long

const int MAXN=500+3;
const int MAXT=1000000000;
int N, M, K, Q;
int y[MAXN*MAXN], x[MAXN*MAXN], t[MAXN*MAXN];
int sum[MAXN][MAXN];

bool judge(int mid)
{
    for(int i=0;i<=N;++i)
        for(int j=0;j<=M;++j)
            sum[i][j]=0;
    for(int i=0;i<Q;++i)
        if(t[i]<=mid)
            sum[y[i]][x[i]]=1;
    for(int i=1;i<=N;++i)
        for(int j=1;j<=M;++j)
        {
            sum[i][j]+=sum[i-1][j]+sum[i][j-1]-sum[i-1][j-1];
            if(i>=K && j>=K && sum[i][j]-sum[i-K][j]-sum[i][j-K]+sum[i-K][j-K]==K*K)
                return true;
        }
    return false;
}

int main()
{
    scanf("%d%d%d%d", &N, &M, &K, &Q);
    for(int i=0;i<Q;++i)
        scanf("%d%d%d", &y[i], &x[i], &t[i]);
    int l=-1, r=MAXT+1;
    while(r-l>1)
    {
        int mid=(l+r)>>1;
        if(judge(mid))
            r=mid;
        else l=mid;
    }
    printf("%d\n", r==MAXT+1?-1:r);

    return 0;
}