poj1050-To the Max(最大子段和)

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#include <iostream>#include <cstring>#include <algorithm>using namespace std;int s[105][105][105];int f[105];int t[105][105];int A[105][105];struct poj1050 {/*[问题描述]:用矩阵内部的所有元素和来衡量一个矩阵的大小,求最大子矩阵的元素和*//*[解题思路]:令s[i][j][row]表示sum(A[i][row],A[i+1][row],...A[j][row])* s[i][j][row]=sum(A[1][row],...,A[j][row])-sum(A[1][row],...,A[i][row])* f[k]表示数组s[i][j]的以第k个元素结尾的最大子段和,则f[k]=max(s[i][j][k],s[i][j][k]+f[k-1])* t[i][j]表示第i行的前j个元素和,则s[i][j][row]=t[row][j]-t[row][i-1]*/int n;void work() {while (cin >> n) {memset(t, 0, sizeof(t));for (int i = 1; i <= n; i++) {for (int j = 1; j <= n; j++) {cin >> A[i][j];t[i][j] = t[i][j - 1] + A[i][j];}}int Max = -10000000;memset(s, 0, sizeof(s));for (int i = 1; i <= n; i++) {for (int j = i; j <= n; j++) {for (int k = 1; k <= n; k++) {s[i][j][k] = t[k][j] - t[k][i - 1];}f[0] = -100000;for (int k = 1; k <= n; k++) {f[k] = max(s[i][j][k], s[i][j][k] + f[k - 1]);Max = max(Max, f[k]);}}}cout << Max << endl;}}};int main(){poj1050 solution;solution.work();return 0;}

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