国产探花免费观看_亚洲丰满少妇自慰呻吟_97日韩有码在线_资源在线日韩欧美_一区二区精品毛片,辰东完美世界有声小说,欢乐颂第一季,yy玄幻小说排行榜完本

首頁 > 學院 > 開發設計 > 正文

POJ1861-Network(Kruskal)

2019-11-11 03:38:04
字體:
來源:轉載
供稿:網友

Network Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 16497 Accepted: 6545 Special Judge Description

Andrew is working as system administrator and is planning to establish a new network in his company. There will be N hubs in the company, they can be connected to each other using cables. Since each worker of the company must have access to the whole network, each hub must be accessible by cables from any other hub (with possibly some intermediate hubs). Since cables of different types are available and shorter ones are cheaper, it is necessary to make such a plan of hub connection, that the maximum length of a single cable is minimal. There is another PRoblem — not each hub can be connected to any other one because of compatibility problems and building geometry limitations. Of course, Andrew will provide you all necessary information about possible hub connections. You are to help Andrew to find the way to connect hubs so that all above conditions are satisfied. Input

The first line of the input contains two integer numbers: N - the number of hubs in the network (2 <= N <= 1000) and M - the number of possible hub connections (1 <= M <= 15000). All hubs are numbered from 1 to N. The following M lines contain information about possible connections - the numbers of two hubs, which can be connected and the cable length required to connect them. Length is a positive integer number that does not exceed 106. There will be no more than one way to connect two hubs. A hub cannot be connected to itself. There will always be at least one way to connect all hubs. Output

Output first the maximum length of a single cable in your hub connection plan (the value you should minimize). Then output your plan: first output P - the number of cables used, then output P pairs of integer numbers - numbers of hubs connected by the corresponding cable. Separate numbers by spaces and/or line breaks. Sample Input

4 6 1 2 1 1 3 1 1 4 2 2 3 1 3 4 1 2 4 1 Sample Output

1 4 1 2 1 3 2 3 3 4

#include<cstdio>#include<iostream>#include<algorithm>#include<stdlib.h>#include<vector>using namespace std;const int maxn=1005;typedef struct edge{ int u,v,w;}edge;edge bian[15005];int pre[maxn];void init(){ for(int i=0;i<maxn;i++) pre[i]=i;}int find(int x){ if(pre[x] == x) return pre[x]; else return pre[x]=find(pre[x]);}void merge(int x,int y){ int fx=find(x),fy=find(y); if(fx != fy) pre[fx]=fy;}int cmp(edge a,edge b){ return a.w <b.w;}int main(){ // freopen("in.txt","r",stdin); int n,m; int t1,t2,t3; while(cin>>n>>m){ for(int i=0;i < m;i++){ cin>>t1>>t2>>t3; bian[i].u=t1,bian[i].v=t2,bian[i].w=t3; } init(); sort(bian,bian+m,cmp); /* for(int i=0;i< m;i++) cout<<bian[i].w<<" "; cout<<endl; */ int rst=n; vector<edge> vii; for(int i=0; i < m && rst>1 ; i++){ int x=bian[i].u,y=bian[i].v,z=bian[i].w; if(find(x)!=find(y)){ merge(x,y); rst --; vii.push_back( bian[i]); } } cout<<vii[ vii.size()-1].w<<endl; cout<<n-1<<endl; for(int i=0;i <vii.size();i++) cout<<vii[i].u<<" "<<vii[i].v<<endl; } return 0;}
發表評論 共有條評論
用戶名: 密碼:
驗證碼: 匿名發表
主站蜘蛛池模板: 札达县| 峡江县| 甘孜县| 临西县| 灯塔市| 南和县| 白朗县| 金华市| 德令哈市| 福泉市| 凌源市| 闻喜县| 南漳县| 巫山县| 黄平县| 丁青县| 伊宁市| 隆林| 成安县| 青海省| 凤山市| 吴忠市| 田阳县| 青冈县| 磐石市| 永平县| 杭州市| 古蔺县| 渭源县| 金华市| 莆田市| 仲巴县| 惠州市| 比如县| 札达县| 孝义市| 德清县| 河源市| 格尔木市| 瑞丽市| 阿克陶县|