UVA 10827 Maximum sum on a torus 可循环的最大连续子矩阵

这题其实是uva 108的拓展，就是给出的矩阵中子矩阵的定义变成可循环的，比如下面这样是一个3*3的子矩阵：

1

-1

0

0

-4

2

3

-2

-3

2

4

1

-1

5

0

3

-2

1

-3

2

-3

2

4

1

-4

只要把矩阵拓展成四倍大，然后处理一下子矩阵不能大于N*N，然后按普通的最大连续矩阵做就可以了。

代码：

/* *   Author:        illuz <iilluzen[at]gmail.com> *   Blog:          http://blog.csdn.net/hcbbt *   File:          uva827.cpp *   Lauguage:      C/C++ *   Create Date:   2013-09-05 16:36:28 *   Descripton:    UVA 10827 Maximum sum on a torus, dp */#include <cstdio>#include <algorithm>using namespace std;#define rep(i, n) for (int i = 0; i < (n); i++)#define repf(i, a, b) for (int i = (a); i <= (b); i++)const int MAXN = 160;int m[MAXN][MAXN], s[MAXN][MAXN], a[MAXN], n;void maxSeq(int* a, int len, int &res) {res = a[0];int sum = 0;for (int i = 0; i < len; i++) {if (sum <= 0)sum = a[i];else sum += a[i];if (res < sum)res = sum;}}int main() {int t;scanf("%d", &t);while (t--) {scanf("%d", &n);repf(i, 1, n) repf(j, 1, n) {scanf("%d", &m[i][j]);m[i + n][j] = m[i][j + n] = m[i + n][j + n] = m[i][j];}repf(i, 1, 2 * n - 1) repf(j, 1, 2 * n - 1)s[i][j] = s[i - 1][j] + m[i][j];int ans = -0xfffffff, res;repf(i, 0, n - 1) repf(j, i + 1, i + n) {rep(l, n) {repf(k, 1, n) a[k] = s[j][k + l] - s[i][k + l];maxSeq(a + 1, n, res);ans = max(ans, res);}}printf("%d\n", ans);}return 0;}