题目链接:​​http://acm.hdu.edu.cn/showproblem.php?pid=5115​​

题意：
有n只狼，每只狼有两种属性，一种攻击力一种附加值，我们没杀一只狼，那么我们受到的伤害值为这只狼的攻击值与它旁边的两只狼的附加值的和，求把所有狼都杀光受到的最小的伤害值。

代码：

#include <stdio.h>  
#include <ctime>  
#include <math.h>  
#include <limits.h>  
#include <complex>  
#include <string>  
#include <functional>  
#include <iterator>  
#include <algorithm>  
#include <vector>  
#include <stack>  
#include <queue>  
#include <set>  
#include <map>  
#include <list>  
#include <bitset>  
#include <sstream>  
#include <iomanip>  
#include <fstream>  
#include <iostream>  
#include <ctime>  
#include <cmath>  
#include <cstring>  
#include <cstdio>  
#include <time.h>  
#include <ctype.h>  
#include <string.h>  
#include <assert.h>  

using namespace std;

int n;
int a[500],b[500],dp[500][500];

int main()
{
    int t, cases = 1;
    scanf("%d",&t);
    while (t--)
    {
        scanf("%d",&n);
        for (int i = 0; i <= n; i++)
            for (int j = 0; j <= n; j++)
                dp[i][j] = 100000000;
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));

        for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
        for (int i = 1; i <= n; i++) scanf("%d", &b[i]);
        for (int i = 1; i <= n; i++) dp[i][i] = a[i] + b[i - 1] + b[i + 1];

        for (int l = 1; l <= n; l++)
        {
            for (int i = 1; i + l <= n; i++)
            {
                int j = i + l;
                dp[i][j] = min(dp[i+1][j] + b[i-1] + b[j+1] + a[i], dp[i][j-1] + a[j] + b[j + 1] + b[i-1]);
                for (int k = i+1; k <= j; k++)
                {
                    dp[i][j] = min(dp[i][j], dp[i][k - 1] + dp[k + 1][j] + a[k] + b[i - 1] + b[j + 1]);
                }
            }
        }
        printf("Case #%d: %d\n", cases++, dp[1][n]);
    }
    return 0;
}