[Prim] POJ 1287

题意：

应该就是装宽带网络，给定节点和有权值的边的数目，问最小的成本是多少。

思路：

Prim模板，第一次

#include <algorithm>#include <cstdio>#include <cstring>#define N 55#define INF 0x7f7f7f7fusing namespace std;int n, m;int edg[ N ][ N ];int dis[ N ];bool vis[ N ];void Prim () {    int ans = 0;    // INIt    memset ( vis, false, sizeof ( vis ) );    vis[ 1 ] = true;    for ( int i = 1; i <= n; ++i ) {        dis[ i ] = edg[ 1 ][ i ];    }    // u = Extract MIN    for ( int i = 1; i <= n; ++i ) {        int x = -1, mi = INF;        for ( int j = 1; j <= n; ++j )            if ( !vis[ j ] && dis[ j ] < mi )                mi = dis[ x = j ];        if ( x == -1 )            break;        //出列后标记        vis[ x ] = true;        ans += mi;        // for each v in G.adj[u]        for ( int j = 1; j <= n; ++j )            // if v in Q and w(u,v) < v.key            // 和dijkstra不一样的地方            if ( !vis[ j ] && dis[ j ] > edg[ x ][ j ] )                dis[ j ] = edg[ x ][ j ];    }    printf ( "%d\n", ans );}int main () {    while ( ~scanf ( "%d", &n ) && n ) {        scanf ( "%d", &m );        if ( !m ) {            printf ( "0\n" );            continue;        }        for ( int i = 1; i <= n; ++i )            for ( int j = 1; j <= n; ++j ) {                if ( i == j )                    edg[ i ][ j ] = 0;                else                    edg[ i ][ j ] = INF;            }        for ( int i = 1; i <= m; ++i ) {            int u, v, w;            scanf ( "%d%d%d", &u, &v, &w );            if ( w < edg[ u ][ v ] )                edg[ u ][ v ] = edg[ v ][ u ] = w;        }        Prim ();    }    return 0;}