hdu 4340

地址：http://acm.hdu.edu.cn/showproblem.php?pid=4340

题意：两个人抢劫一个城市，每个人抢前一个城市相邻的城市的时候只用花费1/2的时间，求最短时间。

mark：树状dp。dp[i][j][k]代表第i个城市由第j个人抢劫，第i个城市所属子树需要完全时间的个数。

代码：

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

const int M = 105;
const int N = 10000000;
int n,s[2][M];
int dp[M][2][2];
bool adj[M][M];

int min(int a, int b) {return a < b ? a : b;}

int dfs(int i, int j, int k, int father = 0)
{
    if(dp[i][j][k]) return dp[i][j][k];
    if(k)
    {
        dp[i][j][k] = s[j][i];
        for(int p = 1; p <= n; p++)
            if(p != father && adj[i][p])
                dp[i][j][k] += min(dfs(p, j, 0, i), dfs(p, !j, 1, i));
        int min1, sum;
        min1 = N;
        sum = 0;
        for(int p = 1; p <= n; p++)
        {
            if(p != father && adj[i][p])
            {
                sum = dfs(p, j, 1, i);
                for(int q = 1; q <= n; q++)
                    if(q != father && q != p && adj[i][q])
                        sum += min(dfs(q, j, 0, i), dfs(q, !j, 1, i));
                min1 = min(min1, sum);
            }
        }
        if(!sum) return dp[i][j][k];
        return dp[i][j][k] = min(dp[i][j][k], min1+s[j][i]/2);
    }
    else
    {
        dp[i][j][k] = s[j][i]/2;
        for(int p = 1; p <= n; p++)
            if(p != father && adj[i][p])
                dp[i][j][k] += min(dfs(p, j, 0, i), dfs(p, !j, 1, i));
        return dp[i][j][k];
    }
}

int main()
{
    int i,j,k;
    while(~scanf("%d", &n))
    {
        for(j = 0; j < 2; j++)
            for(i = 1; i <= n; i++)
                scanf("%d", &s[j][i]);
        memset(adj, 0, sizeof(adj));
        memset(dp, 0, sizeof(dp));
        for(i = 1; i < n; i++)
        {
            scanf("%d %d", &j, &k);
            adj[j][k] = adj[k][j] = 1;
        }
        printf("%d\n", min(dfs(1, 0, 1), dfs(1, 1, 1)));
    }
    return 0;
}