BZOJ1601 [Usaco2008 Oct]灌水[最小生成树]

显然分析可知这个图最后连起来是一个森林，每棵树有一个根再算一个代价。那么这些跟需要连向某一点一个建立水库的代价，且根可以有多个但不能没有，则考虑用超级源点0向所有点连虚边，Prim跑MST即可保证有至少一个根。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 #include<queue>
 #define dbg(x) cerr << #x << " = " << x <<endl
 #define dbg2(x,y) cerr<< #x <<" = "<< x <<"  "<< #y <<" = "<< y <<endl
 using namespace std;
 typedef long long ll;
 typedef double db;
 typedef pair<int,int> pii;
 template<typename T>inline T _min(T A,T B){return A<B?A:B;}
 template<typename T>inline T _max(T A,T B){return A>B?A:B;}
 template<typename T>inline char MIN(T&A,T B){return A>B?(A=B,):;}
 template<typename T>inline char MAX(T&A,T B){return A<B?(A=B,):;}
 template<typename T>inline void _swap(T&A,T&B){A^=B^=A^=B;}
 template<typename T>inline T read(T&x){
     x=;int f=;char c;while(!isdigit(c=getchar()))if(c=='-')f=;
     while(isdigit(c))x=x*+(c&),c=getchar();return f?x=-x:x;
 }
 const int N=+,INF=0x3f3f3f3f;
 int mp[N][N],dis[N],vis[N];
 int n,ans;
 
 int main(){//freopen("test.in","r",stdin);//freopen("test.ans","w",stdout);
     read(n);for(register int i=;i<=n;++i)read(dis[i]);
     for(register int i=;i<=n;++i)for(register int j=;j<=n;++j)read(mp[i][j]);
     for(register int i=,tmp=INF,pt;i<=n;++i,tmp=INF){
         for(register int j=;j<=n;++j)if(!vis[j]&&MIN(tmp,dis[j]))pt=j;
         ans+=tmp;vis[pt]=;
         for(register int j=;j<=n;++j)if(!vis[j])MIN(dis[j],mp[pt][j]);
     }
     return printf("%d\n",ans),;
 }

总结：连虚边向超级源点的套路要记住。