loj 1037(状压dp)

题目链接：http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=25914

思路：dp[state]表示当前状态下要消耗的最小的shots。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define inf 1<<30

int n,val[17],dp[1<<17];
char map[17][17];

int main()
{
    int _case,t=1;
    scanf("%d",&_case);
    while(_case--){
        scanf("%d",&n);
        for(int i=0;i<n;i++)scanf("%d",&val[i]);
        for(int i=0;i<n;i++)scanf("%s",map[i]);
        fill(dp,dp+(1<<n),inf);
        for(int i=0;i<n;i++)dp[1<<i]=val[i];
        for(int state=0;state<(1<<n);state++){
            if(dp[state]==inf)continue;
            for(int i=0;i<n;i++)if(state&(1<<i)){
                for(int j=0;j<n;j++)if(!(state&(1<<j))){
                    int cnt=map[i][j]-'0';
                    if(cnt==0)dp[state|(1<<j)]=min(dp[state|(1<<j)],dp[state]+val[j]);
                    else {
                        cnt=val[j]/cnt+(val[j]%cnt!=0);
                        dp[state|(1<<j)]=min(dp[state|(1<<j)],dp[state]+cnt);
                    }
                }
            }
        }
        printf("Case %d: %d
",t++,dp[(1<<n)-1]);
    }
    return 0;
}