C - Monitor CodeForces - 846D （二维前缀和 + 二分）

发布时间：2020-12-16 07:19:24 所属栏目：百科来源：网络整理

导读：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

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?x_i,?y_i,?t_i?(1?≤?x_i?≤?n,?1?≤?y_i?≤?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?t_i.

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

Output

Input

Output

-1
思路：二分时间或者二分会被毁坏的像素点数量也可以吧，后者虽然没写但是我觉得可以。
　　然后运用二维前缀和，来查找是否存在k*k的像素缺口。
关于二维前缀和?https://blog.csdn.net/yzyyylx/article/details/78298318

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int P = 998244353;
const int inf = 0x3f3f3f3f;
const int maxn = 500030;
struct node{
    ll x,y;
    ll t;
}pos[maxn];
ll n,m,k,q;
ll mp[510][510];
ll a[510][510],b[510][510];
int cmp(struct node & a,struct node & b){
    return a.t<b.t;
}

int check(ll rr){
    memset(a,0,sizeof(a));
    memset(b,sizeof(b));
    for(int i = 1; pos[i].t<= rr&&i <= q ; i++){
        a[pos[i].x][pos[i].y] = 1;
    }
    for(int i = 1; i <= n ; i++){
        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++){
        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;
}

ll ans = -1;
ll Min = 1e10,Max = -1;

int main(){
    memset(mp,-1,sizeof(mp));
    scanf("%lld %lld %lld %lld",&n,&m,&k,&q);
    for(int i = 1; i <= q ; i++){
        scanf("%lld %lld %lld",&pos[i].x,&pos[i].y,&pos[i].t);
        mp[pos[i].x][pos[i].y] = pos[i].t;
        Min = min(Min,pos[i].t);
        Max = max(Max,pos[i].t);
    }
    sort(pos + 1,pos + 1 + q,cmp);
    ll l = Min,r = Max + 1;
    while(r - l > 0){
        ll mid = (r + l)/2;
        if(check(mid)){
            ans = mid;
            r = mid;
        }else{
            l = mid + 1;
        }
    }
    printf("%lldn",ans);
    return 0;
}

View Code

一个从很久以前就开始做的梦。

（编辑：李大同）

【声明】本站内容均来自网络，其相关言论仅代表作者个人观点，不代表本站立场。若无意侵犯到您的权利，请及时与联系站长删除相关内容!