[POJ1050]To the Max (矩阵,最大连续子序列和)

Posted 2021-01-29 lincold

数据弱，暴力过

题意

N^N的矩阵，求最大子矩阵和

思路

悬线？不需要。暴力+前缀和过

代码

//poj1050
//n^4暴力
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#define N 105
#define INF 0x3fffffff
using namespace std;
int a[N][N];
int sum[N];
int ans;
int main()
{
    int n; scanf("%d", &n);
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            scanf("%d", &a[i][j]);
    for (int i = 1; i <= n; i++) {
        memset(sym, 0, sizeof(sum));
        for (int j = i; j <= n; j++) {
            int num = 0;
            for (int k = 1; k <= n; k++) {
                num += sum[k] += a[j][k];
                ans = max(ans, num);
                if (num < 0) num = 0;
            }
        }
    }
    printf("%d
", ans);
    return 0;
}

PS：VS编译坏了，伤心！