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。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 using namespace std; 6 #define inf 1<<30 7 8 int n,val[17],dp[1<<17]; 9 char map[17][17]; 10 11 int main() 12 { 13 int _case,t=1; 14 scanf("%d",&_case); 15 while(_case--){ 16 scanf("%d",&n); 17 for(int i=0;i<n;i++)scanf("%d",&val[i]); 18 for(int i=0;i<n;i++)scanf("%s",map[i]); 19 fill(dp,dp+(1<<n),inf); 20 for(int i=0;i<n;i++)dp[1<<i]=val[i]; 21 for(int state=0;state<(1<<n);state++){ 22 if(dp[state]==inf)continue; 23 for(int i=0;i<n;i++)if(state&(1<<i)){ 24 for(int j=0;j<n;j++)if(!(state&(1<<j))){ 25 int cnt=map[i][j]-'0'; 26 if(cnt==0)dp[state|(1<<j)]=min(dp[state|(1<<j)],dp[state]+val[j]); 27 else { 28 cnt=val[j]/cnt+(val[j]%cnt!=0); 29 dp[state|(1<<j)]=min(dp[state|(1<<j)],dp[state]+cnt); 30 } 31 } 32 } 33 } 34 printf("Case %d: %d\n",t++,dp[(1<<n)-1]); 35 } 36 return 0; 37 }
