hdu 5115 区间dp

dp[i][j]代表消灭这个区间中的狼消耗的最小代价

转移过程中枚举每一个点作为最后一个杀死点的最小代价

代码如下：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#define MAX 207

using namespace std;

int dp[MAX][MAX];
int a[MAX];
int b[MAX];

int main ( )
{
    int t,n;
    scanf ( "%d" , &t );
    int c = 1;
    while ( t-- )
    {
        scanf ( "%d" , &n );
        memset ( dp , 0x3f , sizeof ( dp ) );
        b[0] = 0 , b[n+1] = 0;
        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 i = 1 ; i < n ; i++ )
            for ( int j = 1 ; j <= n-i ; j++ )
                for ( int k = j ; k <= j+i ; k++ )
                if ( k == j ) 
                    dp[j][j+i] =
                min (dp[j+1][j+i] + a[k] + b[j-1] + b[j+i+1] , dp[j][j+i]) ;
                else if ( k == j+i )
                    dp[j][j+i] =
                min ( dp[j][j+i], dp[j][j+i-1] + a[k] + b[j+i+1]+b[j-1] );
                else 
                dp[j][j+i] =
        min ( dp[j][j+i], dp[j][k-1] +b[j-1] + a[k] + b[j+i+1] +dp[k+1][j+i]);
        printf ( "Case #%d: " , c++ );
        printf ( "%d\n" , dp[1][n] );  
    }
}