对于 N=8 的矩阵，如下序列都是合法的：
M2,1,M2,2,M2,3,M2,4,M2,5,M2,6,M2,7,M2,8.
M2,2,M2,3,M2,4.
M2,6,M2,7,M2,8,M2,1,M2,2.
M4,3,M5,3,M6,3,M7,3.
M1,2,M2,3,M3,4,M4,5.
M2,4,M3,3,M4,2,M5,1.
M3,3,M4,2,M5,1,M1,5.
M5,6.
一个元素不可取多次，取的必须是连续的一段。
可以什么都不取（即答案为 0）。
标签：def from 技术 for 结果 pre i++ span stdin
1
4
8 6 6 1
-3 4 0 5
4 2 1 9
1 -9 9 -2
24
样例解释：选取序列 M3,4,M4,3,M1,2。
#include "bits/stdc++.h" using namespace std; const int inf = 0x3f3f3f3f; const int maxn = 1100; int a[maxn][maxn], b[maxn][maxn]; int n; int f1[maxn], f2[maxn]; int ans; int main() { freopen("input.txt", "r", stdin); int _; scanf("%d", &_); while (_--) { scanf("%d", &n); ans = -inf; for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) { scanf("%d", &a[i][j]); b[i][j] = -a[i][j]; } } int sum, max1, max2; for (int i = 1; i <= n; i++) { sum = 0, max1 = -inf, max2 = -inf;//max2为取反后最大子串和 for (int j = 1; j <= n; j++) { sum += a[i][j]; f1[j] = 0; f2[j] = 0; } for (int j = 1; j <= n; j++) { if (f1[j - 1] >= 0) { f1[j] = f1[j - 1] + a[i][j]; } else f1[j] = a[i][j]; if (f2[j - 1] >= 0) { f2[j] = f2[j - 1] + b[i][j]; } else f2[j] = b[i][j]; max1 = max(max1, f1[j]); max2 = max(max2, f2[j]); } ans = max(ans, max(sum + max2, max1)); sum = 0, max1 = -inf, max2 = -inf; for (int j = 1; j <= n; j++) { sum += a[j][i]; f1[j] = 0; f2[j] = 0; } for (int j = 1; j <= n; j++) { if (f1[j - 1] >= 0) { f1[j] = f1[j - 1] + a[j][i]; } else f1[j] = a[j][i]; if (f2[j - 1] >= 0) { f2[j] = f2[j - 1] + b[j][i]; } else f2[j] = b[j][i]; max1 = max(max1, f1[j]); max2 = max(max2, f2[j]); } ans = max(ans, max(sum + max2, max1)); sum = 0, max1 = -inf, max2 = -inf; for (int j = 1; j <= n; j++) { sum += a[j][(j + i - 1) % n + 1]; f1[j] = 0; f2[j] = 0; } for (int j = 1; j <= n; j++) { if (f1[j - 1] >= 0) { f1[j] = f1[j - 1] + a[j][(j + i - 1) % n + 1]; } else f1[j] = a[j][(j + i - 1) % n + 1]; if (f2[j - 1] >= 0) { f2[j] = f2[j - 1] + b[j][(j + i - 1) % n + 1]; } else f2[j] = b[j][(j + i - 1) % n + 1]; max1 = max(max1, f1[j]); max2 = max(max2, f2[j]); } ans = max(ans, max(sum + max2, max1)); sum = 0, max1 = -inf, max2 = -inf; int t; for (int j = 1; j <= n; j++) { t = (n - j - 1 + n) % n + 1; sum += a[t][(j - i - 1 + n) % n + 1]; f1[j] = 0; f2[j] = 0; } for (int j = 1; j <= n; j++) { t = (n - j - 1 + n) % n + 1; if (f1[j - 1] >= 0) { f1[j] = f1[j - 1] + a[t][(j - i - 1 + n) % n + 1]; } else f1[j] = a[t][(j - i - 1 + n) % n + 1]; if (f2[j - 1] >= 0) { f2[j] = f2[j - 1] + b[t][(j - i - 1 + n) % n + 1]; } else f2[j] = b[t][(j - i - 1 + n) % n + 1]; max1 = max(max1, f1[j]); max2 = max(max2, f2[j]); } ans = max(ans, max(sum + max2, max1)); } if (ans < 0) ans = 0; printf("%d\n", ans); } return 0; }
UPC3463: Maximal-sum Subsequence
标签：def from 技术 for 结果 pre i++ span stdin
原文地址：https://www.cnblogs.com/albert-biu/p/10339716.html
