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Description
Input
Output
Sample Input
Sample Output
Hint
Hint Huge input, scanf is recommended.
最小生成树的题目,因为点比较少,可以直接用普里姆算法。
#include <iostream> #include <algorithm> #include <cstring> #include <cstdio> #include <cctype> #include <cstdlib> #include <stack> #include <cmath> #include <string> #include <queue> #define INF 0xffffff using namespace std; const int maxn = 101; typedef long long ll ; int n, m, x, y, v; int G[maxn][maxn], d[maxn]; bool vis[maxn]; int prim(){ int pt, ret, Min; memset(vis,false,sizeof(vis)); memset(d,0x7f,sizeof(d)); pt = 1; vis[1] = true; ret = 0; while( true ){ for(int i=1; i<=n; i++) if( !vis[i] && G[pt][i] && d[i]>G[pt][i] ) d[i] = G[pt][i]; pt = -1; Min = INF ; for(int i=1; i<=n; i++){ if( !vis[i] && Min > d[i] ){ Min = d[i]; pt = i; } } if( pt == -1 ) break; ret += Min; vis[pt] = true; } return ret; } int main(){ while( ~scanf("%d",&n) && n){ m = n * ( n - 1 ) / 2 ; memset(G,0,sizeof(G)); while( m -- ){ scanf("%d %d %d",&x,&y,&v); G[x][y] = G[y][x] = v ; } printf("%d\n",prim()); } return 0 ; }
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原文地址:http://www.cnblogs.com/Asimple/p/5699484.html