牛客

https://ac.nowcoder.com/acm/contest/700/I
二维RMQ，贴个板子，注意爆内存，用char就可以了，char也可以存负数。
然后二分枚举对角线长度，理由很简单。
矩阵变大，极值只会变大不会变小。满足单调性。

#include<bits/stdc++.h>
using namespace std;
#define MAXN 501

int n,m,g;
int rec[MAXN][MAXN];
char dp[MAXN][MAXN][11][11];
char dp1[MAXN][MAXN][11][11];
inline int maxm(int a,int b,int c,int d) {
    if(a<b)
        a=b;
    if(a<c)
        a=c;
    if(a<d)
        a=d;
    return a;
}
inline int minm(int a,int b,int c,int d) {
    if(b<a)
        a=b;
    if(c<a)
        a=c;
    if(d<a)
        a=d;
    return a;
}
void st() {
    for(int k=0; (1<<k)<=n; k++)
        for(int l=0; (1<<l)<=m; l++)
            for(int i=1; i+(1<<k)-1<=n; i++)
                for(int j=1; j+(1<<l)-1<=m; j++) {
                    if(!k&&!l) {
                        dp1[i][j][k][l]=dp[i][j][k][l]=rec[i][j];
                    } else if(k==0) {
                        dp[i][j][k][l]=max(dp[i][j][k][l-1],dp[i][j+(1<<(l-1))][k][l-1]);
                        dp1[i][j][k][l]=min(dp1[i][j][k][l-1],dp1[i][j+(1<<(l-1))][k][l-1]);
                    } else if(l==0) {
                        dp[i][j][k][l]=max(dp[i][j][k-1][l],dp[i+(1<<(k-1))][j][k-1][l]);
                        dp1[i][j][k][l]=min(dp1[i][j][k-1][l],dp1[i+(1<<(k-1))][j][k-1][l]);
                    } else {
                        dp[i][j][k][l]=maxm(dp[i][j][k-1][l-1],dp[i+(1<<(k-1))][j][k-1][l-1],
                                            dp[i][j+(1<<(l-1))][k-1][l-1],dp[i+(1<<(k-1))][j+(1<<(l-1))][k-1][l-1]);
                        dp1[i][j][k][l]=minm(dp1[i][j][k-1][l-1],dp1[i+(1<<(k-1))][j][k-1][l-1],
                                             dp1[i][j+(1<<(l-1))][k-1][l-1],dp1[i+(1<<(k-1))][j+(1<<(l-1))][k-1][l-1]);
                    }
                    //printf("dp[%d][%d][%d][%d]=%d
",i,j,k,l,dp[i][j][k][l]);
                }
}
int rmq2dmax(int x,int y,int x1,int y1) {
    int k=0;
    while((x1-x+1)>=(1<<k))
        k++;
    k--;
    int l=0;
    while((y1-y+1)>=(1<<l))
        l++;
    l--;
    return maxm(dp[x][y][k][l],dp[x1-(1<<k)+1][y][k][l],
                dp[x][y1-(1<<l)+1][k][l],dp[x1-(1<<k)+1][y1-(1<<l)+1][k][l]);
}

int rmq2dmin(int x,int y,int x1,int y1) {
    int k=0;
    while((x1-x+1)>=(1<<k))
        k++;
    k--;
    int l=0;
    while((y1-y+1)>=(1<<l))
        l++;
    l--;
    return minm(dp1[x][y][k][l],dp1[x1-(1<<k)+1][y][k][l],
                dp1[x][y1-(1<<l)+1][k][l],dp1[x1-(1<<k)+1][y1-(1<<l)+1][k][l]);
}

bool check(int len,int &suc) {
    for(int i=1; i<=n; i++) {
        if(i+len-1>n)
            break;
        for(int j=1; j<=m; j++) {
            if(j+len-1>m)
                break;
            int t=rmq2dmax(i,j,i+len-1,j+len-1)-rmq2dmin(i,j,i+len-1,j+len-1);
            if(t<=g) {
                suc=1;
                return 1;
            }
        }
    }
    return 0;
}

int main() {
    scanf("%d%d%d",&n,&m,&g);
    for(int i=1; i<=n; i++) {
        for(int j=1; j<=m; j++) {
            scanf("%d",&rec[i][j]);
        }
    }
    st();


    int l=1,r=min(n,m);
    int ans=-1;

    while(1) {
        int len=(l+r)>>1;
        int suc=0;

        if(l==len){
            check(r,suc);
            if(suc){
                ans=r;
            }
            else{
                ans=l;
            }
            break;
        }

        check(len,suc);

        if(suc) {
            l=len;
        } else {
            r=len-1;
        }
    }

    printf("%d
",ans);
}