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,1):0;}
template<typename T>inline char MAX(T&A,T B){return A<B?(A=B,1):0;}
template<typename T>inline void _swap(T&A,T&B){A^=B^=A^=B;}
template<typename T>inline T read(T&x){
    x=0;int f=0;char c;while(!isdigit(c=getchar()))if(c=='-')f=1;
    while(isdigit(c))x=x*10+(c&15),c=getchar();return f?x=-x:x;
}
const int N=300+7,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=1;i<=n;++i)read(dis[i]);
    for(register int i=1;i<=n;++i)for(register int j=1;j<=n;++j)read(mp[i][j]);
    for(register int i=1,tmp=INF,pt;i<=n;++i,tmp=INF){
        for(register int j=1;j<=n;++j)if(!vis[j]&&MIN(tmp,dis[j]))pt=j;
        ans+=tmp;vis[pt]=1;
        for(register int j=1;j<=n;++j)if(!vis[j])MIN(dis[j],mp[pt][j]);
    }
    return printf("%d
",ans),0;
}

总结：连虚边向超级源点的套路要记住。