NYOJ 布线问题 prim

 //普通邻接矩阵  时间复杂度 O(v*e)#include <iostream>#include <stdlib.h>#include <stdio.h>#include <string.h>#define INF 999999int v;int map[505][505];int lowcost[505];int visit[1000];int prim( int start ){int min, choose, ans;ans = 0;for( int i = 1; i <= v; i++ ){lowcost[i] = map[start][i];}lowcost[start] = 0;visit[start] = 1;for( int i = 1; i <= v-1; i++ ){min = INF;for( int j = 1; j <= v; j++ ){if( !visit[j] && lowcost[j] >= 0 && lowcost[j] < min ){min = lowcost[j];choose = j;}}ans += lowcost[choose];visit[choose] = 1;for( int j = 1; j <= v; j++ ){if( !visit[j] && map[choose][j] < lowcost[j] )lowcost[j] = map[choose][j];}}return ans;}int main( ){int T, e, min, a, b, w, num;scanf( "%d", &T );while( T-- ){memset( map, 0, sizeof(map) );memset( visit, 0, sizeof(visit) );memset( lowcost, -1, sizeof(lowcost) );scanf( "%d%d", &v, &e );for( int i = 0; i < e; i++ ){scanf( "%d %d %d\n", &a, &b, &w );if( map[a][b] ){if( map[a][b] > w ){map[a][b] = w;map[b][a] = w;}}else{map[a][b] = w;map[b][a] = w;}}min = INF;for( int i = 0; i < v; i++ ){scanf( "%d", &num );if( num < min )min = num;}printf( "%d\n", min+prim( 1 ) );}return 0;}                                       

优先队列+邻接矩阵  优先队列 时间复杂度  O( nlog(n) )

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <queue>#define MAX 505#define INF 9999999#define CLR(arr,val) memset(arr,val,sizeof(arr))using namespace std;typedef struct a{int dis, v;friend bool operator < ( const struct a &n1, const struct a &n2 ){return n1.dis > n2.dis;}}node;int n;int map[MAX][MAX];int vis[MAX];int lowcost[MAX];inline void init(){    for( int i = 0; i < MAX; i++ )    {        lowcost[i] = INF;        for( int j = 0; j < MAX; j++ )            map[i][j] = INF;    }    CLR( vis, 0 );  };int prim( int start ){int sum = 0;node pre, cur;priority_queue<node> Q;lowcost[start] = 0;pre.dis = 0;pre.v = start;Q.push( pre );while( !Q.empty() ){pre = Q.top();Q.pop();if( vis[pre.v] )continue;sum += pre.dis;vis[pre.v] = 1;for( int i = 1; i <= n; i++ ){if( !vis[i] &&  map[pre.v][i] < lowcost[i] ){lowcost[i] = map[pre.v][i];cur.dis = lowcost[i];cur.v = i;Q.push( cur );}}}return sum;}int main(){    int T, e, a, b, w, num, min;    scanf( "%d", &T );      while( T-- )      {      init();        scanf( "%d%d", &n, &e );          for( int i = 0; i < e; i++ )          {              scanf( "%d %d %d\n", &a, &b, &w );              if( map[a][b] )              {                  if( map[a][b] > w )                  {                      map[a][b] = w;                      map[b][a] = w;                  }              }              else              {                  map[a][b] = w;                  map[b][a] = w;              }          }          min = INF;          for( int i = 0; i < n; i++ )          {              scanf( "%d", &num );              if( num < min )                  min = num;          }         printf( "%d\n", prim( 1 ) + min );    }return 0;}