题目链接：https://cn.vjudge.net/problem/HihoCoder-1634
题目意思：可以让矩阵里的某一个数变成p，或者不修改。求最大子矩阵最小，输出最小值。
思路：请看下图

代码：

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <algorithm>
#include <stdlib.h>
#include <math.h>using namespace std;
typedef long long int LL;
int max(int a,int b,int c,int d)
{return max(max(a,b),max(c,d));
}
const int maxn=155;
const int INF=2e9+1e8;
int n,m,p;
int a[maxn][maxn],dp[maxn][maxn],ma[maxn][maxn],L[maxn],R[maxn],U[maxn],D[maxn];
void init()
{memset(L, 0x88, sizeof(L));memset(R, 0x88, sizeof(R));memset(U, 0x88, sizeof(U));memset(D, 0x88, sizeof(D));
}
void deal()
{int tmp;//  xiamemset(dp, 0, sizeof(dp));memset(ma, 0x88, sizeof(ma));tmp = -INF;for (int i = 1; i <= n; i++){for (int l = 1; l <= m; l++){int sum = 0;for (int r = l; r <= m; r++){sum += a[i][r];dp[l][r] += sum;ma[l][r] = max(ma[l][r], dp[l][r]);if (dp[l][r] < 0)dp[l][r] = 0;tmp = max(tmp, ma[l][r]);}}D[i] = tmp;}// zuomemset(dp, 0, sizeof(dp));memset(ma, 0x88, sizeof(ma));tmp = -INF;for (int i = m; i>0; i--){for (int l = 1; l <= n; l++){int sum = 0;for (int r = l; r <= n; r++){sum += a[r][i];dp[l][r] += sum;ma[l][r] = max(ma[l][r], dp[l][r]);if (dp[l][r] < 0)dp[l][r] = 0;tmp = max(tmp, ma[l][r]);}}L[i] = tmp;}// shangmemset(dp, 0, sizeof(dp));memset(ma, 0x88, sizeof(ma));tmp = -INF;for (int i = n; i>0; i--){for (int l = 1; l <= m; l++){int sum = 0;for (int r = l; r <= m; r++){sum += a[i][r];dp[l][r] += sum;ma[l][r] = max(ma[l][r], dp[l][r]);if (dp[l][r] < 0)dp[l][r] = 0;tmp = max(tmp, ma[l][r]);}}U[i] = tmp;}// youmemset(dp, 0, sizeof(dp));memset(ma, 0x88, sizeof(ma));tmp = -INF;for (int i = 1; i <= m; i++){for (int l = 1; l <= n; l++){int sum = 0;for (int r = l; r <= n; r++){sum += a[r][i];dp[l][r] += sum;ma[l][r] = max(ma[l][r], dp[l][r]);if (dp[l][r] < 0)dp[l][r] = 0;tmp = max(tmp, ma[l][r]);}}R[i] = tmp;}
}int main()
{while(scanf("%d%d%d",&n,&m,&p)+1){init();for(int i=1;i<=n;i++){for(int j=1;j<=m;j++)scanf("%d",&a[i][j]);}deal();int ans=D[n];for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){int tmp=max(D[i-1],U[i+1],L[j+1],R[j-1]);tmp=max(tmp,D[n]-a[i][j]+p);ans=min(ans,tmp);}}printf("%d\n",ans);}return 0;
}/*
3 3 -10
-100 4 4
4 -10 4
4 4 1
3 3 -1
-2 -2 -2
-2 -2 -2
-2 -2 -2*/