- dp[S]表示已经消灭目标集合为S的最少射击数
- dp[0]=0
- dp[S]=min( dp[S'] + min( health[i] , health[i]/demage[j][i] ) ) 其中S-S'={i}，j∈S'
1 #include<cstdio>
2 #include<cstring>
3 #include<algorithm>
4 using namespace std;
5 #define INF (1<<29)
6 int d[1<<15],a[15][15],b[15];
7 int main(){
8 int t,n;
9 scanf("%d",&t);
10 for(int cse=1; cse<=t; ++cse){
11 scanf("%d",&n);
12 for(int i=0; i<n; ++i) scanf("%d",b+i);
13 for(int i=0; i<n; ++i){
14 for(int j=0; j<n; ++j) scanf("%1d",&a[i][j]);
15 }
16 for(int i=1; i<(1<<n); ++i) d[i]=INF;
17 for(int i=1; i<(1<<n); ++i){
18 for(int j=0; j<n; ++j){
19 if(((i>>j)&1)==0) continue;
20 int time=b[j];
21 for(int k=0; k<n; ++k){
22 if(j==k || ((i>>k)&1)==0 || a[k][j]==0) continue;
23 time=min(time,b[j]/a[k][j]+(b[j]%a[k][j]!=0));
24 }
25 d[i]=min(d[i],d[i^(1<<j)]+time);
26 }
27
28 }
29 printf("Case %d: %d\n",cse,d[(1<<n)-1]);
30 }
31 return 0;
32 }
