POJ 1861 Network //Kruscal算法

来源：互联网发布：量化投资数据挖掘编辑：程序博客网时间：2024/05/01 23:58

题目描述

POJ1861 Network

解题思路

题目大意:
给出顶点数,边数,再输入若干条边,求出若干条边满足条件:
1)这几条边中长度最大的长度尽量小;
2)这若干条边可保证图的连通性;
3)在1,2的基础上,求个MST
Kruscal算法求出的生成树满足一个特性:这个生成树的最长边是尽量短的.

参考代码

#include <iostream>#include <algorithm>#include <cstdio>#define MAX_N 1010#define MAX_E 15010using namespace std;int par[MAX_N];int rank[MAX_N];struct edge{    int u,v,cost;}es[MAX_E];int V,E;bool cmp(const edge& e1,const edge& e2){    return e1.cost < e2.cost;}void init(){    for (int i = 0;i < V;++i){        par[i] = i;        rank[i] = 0;    }}int find(int x){    if (par[x] == x)    return x;    else    return par[x] = find(par[x]);}void unite(int x,int y){    x = find(x);    y = find(y);    if (x == y) return ;    if (rank[x] < rank[y])  par[x] = y;    else{        par[y] = x;        if (rank[x] == rank[y]) rank[x]++;    }}bool same(int x,int y){    return (find(x) == find(y));}void Kruskal(){    sort(es,es+E,cmp);    init();    int res = 0,opr = 0,i;    for (i = 0;i < E;++i){        edge e = es[i];        if (!same(e.u,e.v)){            unite(e.u,e.v);            res += e.cost;            opr++;        }        if(opr == V-1)  break;    }    printf("%d\n%d\n",es[i].cost,i+1);    for (int j = 0;j <= i;++j)        printf("%d %d\n",es[j].u,es[j].v);}int main(){    scanf("%d %d",&V,&E);    for (int i = 0;i < E;++i){        scanf("%d %d %d",&es[i].u,&es[i].v,&es[i].cost);    }    Kruskal();    return 0;}

0 0