poj1251

来源：互联网发布：win7 apache 编辑：程序博客网时间：2024/06/17 05:04

这道题的题意太明显了，求最小生成树。先面关于图的数字说明，这个图最适合用kruscal来求最小生成树。

poj 1251 Jungle Roads 代码：

#include<iostream>#include<fstream>#include<algorithm>#include<string.h>using namespace std;int parent[26];int n;//vertex numint edge_num;struct Edge{int v1;int v2;int weight;}edges[100];int cmp(const void *edge1,const void *edge2){Edge *e1=(Edge *)edge1;Edge *e2=(Edge *)edge2;return e1->weight-e2->weight;}int find(int v){while(parent[v]>=0)v=parent[v];return v;}void Union(int r1,int r2)   {if(parent[r1]<=parent[r2]){parent[r1]+=parent[r2];parent[r2]=r1;}else{parent[r2]+=parent[r1];parent[r1]=r2;}}int main(){freopen("input.txt","r",stdin);while(scanf("%d\n",&n)!=EOF&&n!=0){edge_num=0;memset(parent,-1,sizeof(int)*n);char ch;int v1,v2;int k;for(int t=0;t<n-1;t++){scanf("%c %d",&ch,&k);v1=ch-'A';for(int i=0;i<k;i++){int w;scanf(" %c %d",&ch,&w);edges[edge_num].v1=v1;edges[edge_num].v2=ch-'A';edges[edge_num++].weight=w;}scanf("\n");}qsort(edges,edge_num,sizeof(Edge),cmp);int output=0;int count=0;for(int i=0;i<edge_num;i++){if(count==n-1) //successedbreak;int r1=find(edges[i].v1);int r2=find(edges[i].v2);if(r1==r2)continue;Union(r1,r2);output+=edges[i].weight;count++;}/*for(int i=0;i<edge_num;i++){cout<<edges[i].v1<<" "<<edges[i].v2<<" "<<edges[i].weight<<endl;}*/        printf("%d\n",output);}return 0;}