HihoCoder 1634 Puzzle Game（最大子矩阵和）题解

题意：给一个n*m的矩阵，你只能选择一个格子把这个格子的数换成p（也可以一个都不换），问最大子矩阵和最小可能是多少？

思路：

思路就是上面这个思路，这里简单讲一下怎么n^3求最大子矩阵和：枚举两行（或者两列），然后把每一列之和看做一个数字，这样二维就变成了一维，我们可以直接求最大子串和的方法。初始一个ret为0，然后从左往右加，如果ret<0，那么把ret初始化0，负数作为初始值肯定比重新开始小，然后找出ret的最大值就是最大子矩阵和。

代码：

#include<cstdio>
#include<cstring>
typedef long long ll;
using namespace std;
const int maxn = 150 + 10;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
int num[maxn][maxn], sum[maxn][maxn];
int up[maxn], down[maxn], left[maxn], right[maxn];
int n, m, p, xx1, yy1, xx2, yy2;
int min(int x, int y){
    return x < y? x : y;
}
int max(int x, int y){
    return x > y? x : y;
}
int mat(int x1, int y1, int x2, int y2){
    return sum[x2][y2] - sum[x1 - 1][y2] - sum[x2][y1 - 1] + sum[x1 - 1][y1 - 1];
}
int main(){
    while(~scanf("%d%d%d", &n, &m, &p)){
        memset(sum, 0, sizeof(sum));
        for(int i = 1; i <= n; i++){
            for(int j = 1; j <= m; j++){
                scanf("%d", &num[i][j]);
                sum[i][j] = num[i][j] + sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1];
            }
        }

        //上下左右四个矩阵预处理
        memset(up, -INF ,sizeof(up));
        for(int i = 1; i <= n; i++){
            int tmp = -INF;
            for(int j = 1; j <= i; j++){
                int ret = 0;
                for(int k = 1; k <= m; k++){
                    ret += mat(j, k, i, k);
                    tmp = max(tmp, ret);
                    if(ret < 0) ret = 0;
                }
            }
            up[i] = max(up[i - 1], tmp);
        }
        memset(down, -INF, sizeof(down));
        for(int i = n; i >= 1; i--){
            int tmp = -INF;
            for(int j = i; j <= n; j++){
                int ret = 0;
                for(int k = 1; k <= m; k++){
                    ret += mat(i, k, j, k);
                    tmp = max(tmp, ret);
                    if(ret < 0) ret = 0;
                }
            }
            down[i] = max(down[i + 1], tmp);
        }
        memset(left, -INF, sizeof(left));
        for(int i = 1; i <= m; i++){
            int tmp = -INF;
            for(int j = 1; j <= i; j++){
                int ret = 0;
                for(int k = 1; k <= n; k++){
                    ret += mat(k, j, k, i);
                    tmp = max(ret, tmp);
                    if(ret < 0) ret = 0;
                }
            }
            left[i] = max(left[i - 1], tmp);
        }
        memset(right, -INF, sizeof(right));
        for(int i = m; i >= 1; i--){
            int tmp = -INF;
            for(int j = i; j <= m; j++){
                int ret = 0;
                for(int k = 1; k <= n; k++){
                    ret += mat(k, i, k, j);
                    tmp = max(ret, tmp);
                    if(ret < 0) ret = 0;
                }
            }
            right[i] = max(right[i + 1], tmp);
        }

        int Max = -INF;
        for(int i = 1; i <= n; i++){
            for(int j = 1; j <= i; j++){
                int ret = 0, start = 1;;
                for(int k = 1; k <= m; k++){
                    ret += mat(j, k, i, k);
                    if(ret > Max){
                        Max = ret;
                        xx1 = j, yy1 = start, xx2 = i, yy2 = k;
                    }
                    if(ret < 0) ret = 0, start = k + 1;
                }
            }
        }
        int ans = Max;
        for(int i = xx1; i <= xx2; i++){
            for(int j = yy1; j <= yy2; j++){
                if(p > num[i][j]) continue;
                ans = min(ans , max(Max - num[i][j] + p, max(up[i - 1], max(down[i + 1], max(left[j - 1], right[j + 1])))));
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}

原文地址：https://www.cnblogs.com/KirinSB/p/10301669.html