[动态规划] Sum游戏 ( Game of Sum, Uva 10891 )

抓住状态转移方程即可 : 从子序列 i j 中取最大 = i + 从子序列i+1,j中取最大 或 j + 从子序列i,j-1中取最大

#include <algorithm>#include <cstring>#include <cstdio>using namespace std;const int maxn = 100+10;int S[maxn], A[maxn], d[maxn][maxn], vis[maxn][maxn], n;int dp(int i, int j){if (vis[i][j]) return d[i][j];vis[i][j] = 1;if (i == j) {d[i][j] = A[i];return d[i][j];}d[i][j] = max(A[i] + S[i+1,j] – dp(i + 1,j), A[j] + S[i,j – 1] – dp(i,j – 1));return d[i][j];}int main(){while (scanf("%d", &n) && n){S[0] = 0;for (int i = 1; i <= n;i++){scanf("%d", &A[i]);S[i] = S[i – 1] + A[i];}memset(vis, 0, sizeof(vis));printf("%d\n", 2*dp(1, n)-S[n]);}return 0;}

，只有不快的斧，没有劈不开的柴。