nyoj737 石子合并（一） 区间DP

    dp[x][y]表示合并[x, y]区间的石子的最小花费，将区间长度递增枚举即可。

  AC代码：

#include<cstdio>#include<algorithm>using namespace std;const int inf = 1 << 30;const int maxn = 200 + 5;int dp[maxn][maxn], a[maxn], sum[maxn];int solve(int n){for(int i = 1; i < n; ++i) dp[i][i] = 0;for(int i = 2; i <= n; ++i)for(int j = 1; j <= n - i + 1; ++j){int x = j, y = j + i - 1;dp[x][y] = inf;for(int k = j; k < y; ++k){dp[x][y] = min((dp[x][k] + dp[k + 1][y]) + sum[y] - sum[x - 1], dp[x][y]);} }return dp[1][n];}int main(){int n;while(scanf("%d", &n) == 1){sum[0] = 0;for(int i = 1; i <= n; ++i) {scanf("%d", &a[i]);sum[i] = sum[i - 1] + a[i];}printf("%d\n", solve(n));}return 0;}

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