区间dp->石子合并(洛谷1775)

https://www.luogu.com.cn/problem/P1775

题意：合并石子，每次合并的代价为相邻两堆的重量，求最小代价。

分析：区间dp，枚举所有的状态为下一次的区间长度计算需要的子集做准备，初始dp[i][i] = 0,前缀和计算合成代价。

void solve(){
    int n;
    cin >> n;

    vector<vector<long long>> dp(n + 1, vector<long long> (n + 1, 0x3f3f3f3f));
    vector<long long> pref(n + 1);
    for (int i = 1; i <= n; ++i){
        cin >> pref[i];
        pref[i] += pref[i - 1];
        dp[i][i] = 0;
    }

    for (int len = 2; len <= n; ++len){
        for (int i = 1; i + len - 1 <= n; ++i){
            for (int k = i, j = i + len - 1; k < j; ++k){
                dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j] + pref[j] - pref[i - 1]);
            }
        }
    }

    cout << dp[1][n] << '\n';
}