bzoj1261

区间dp

这道题就是把区间dp套了一件树的外衣，看见这种构造二叉搜索树的题，就可以用区间dp解决，如果优化的话似乎可以用决策单调性？

dp[l][r]表示l->r这些节点构建一颗二叉搜索树的最小频率，那么我们用记忆化搜索转移，dp[l][r]=dp[l][i-1]+dp[i+1][r]+k*(sum[r]-sum[l-1])+a[i]*c) i=l->r

这里枚举的i是指以i为根的树，那么dp式里的各个变量也很好理解，注意当l>r时dp[l][r]=0

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int N = 35;
const double inf = 1e100;
int n;
double k, c;
double dp[N][N], a[N], sum[N];
double dfs(int l, int r)
{
    if(l > r) return dp[l][r] = 0.0;
    if(dp[l][r] != inf) return dp[l][r];
    for(int i = l; i <= r; ++i) dp[l][r] = min(dp[l][r], dfs(l, i - 1) + dfs(i + 1, r) + k * (sum[r] - sum[l - 1]) + c * a[i]);
    return dp[l][r];
}
int main()
{
    cin >> n >> k >> c;
    double S = 0;
    for(int i = 1; i <= n; ++i) scanf("%lf", &a[i]), S += a[i];
    for(int i = 1; i <= n; ++i) a[i] /= S, sum[i] = sum[i - 1] + a[i];
    for(int i = 1; i <= n; ++i)
        for(int j = 1; j <= n; ++j) dp[i][j] = inf;
    printf("%.3f\n", dfs(1, n));
    return 0;
}

View Code

 

转载于:https://www.cnblogs.com/19992147orz/p/7506977.html