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UVA - 507 Jill Rides Again（连续子序列最大和）升级版

题目大意：给出 n*n 的矩阵，找每隔数字之和最大的子矩阵，输出最大和。
解题思路：枚举矩阵左上和右下的坐标，分别合并子矩阵的每列，使得二维转化为一维，然后利用连续子序列最大和去做就行。

#include<iostream>#include<stdio.h>#include<algorithm>#include<cmath>#include<string.h>const int INF = 0x3f3f3f3f;const int NINF = -INF -1;using namespace std;int val[110][110];int sum[110];int N;int main() {    while (scanf("%d", &N) != EOF) {            for (int i = 0; i < N; i++)            for (int j = 0; j < N; j++)                scanf("%d", &val[i][j]);        int maxn = NINF;        for (int i = 0; i < N; i++) {            for (int j = 0; j < N; j++) {                memset(sum, 0, sizeof(sum));                for (int k = i; k < N; k++) {                    int tmp = 0;                    for (int l = j; l < N; l++) {                        sum[l] += val[k][l];                        if (tmp < 0) {                            tmp = 0;                        }                        tmp += sum[l];                        if (tmp > maxn) {                            maxn = tmp;                        }                    }                }            }        }        printf("%d\n", maxn);    }return 0;}

O(n4)可以过，time(ms):30
试着优化，删了一层 for，time(ms):0

#include<iostream>#include<stdio.h>#include<algorithm>#include<cmath>#include<string.h>const int INF = 0x3f3f3f3f;const int NINF = -INF -1;using namespace std;int val[110][110];int sum[110];int N;int main() {    scanf("%d", &N);    for (int i = 0; i < N; i++)        for (int j = 0; j < N; j++)            scanf("%d", &val[i][j]);    int maxn = NINF;    for (int i = 0; i < N; i++) {        memset(sum, 0, sizeof(sum));        for (int j = i; j < N; j++) {            int tmp = 0;            for (int k = 0; k < N; k++) {                sum[k] += val[j][k];                if (tmp < 0) {                    tmp = 0;                }                tmp += sum[k];                if (tmp > maxn) {                    maxn = tmp;                }            }        }    }    printf("%d\n", maxn);return 0;}

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