POJ 1861：Network（最小產生樹&&kruskal）

標籤：poj

Network

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 13266		Accepted: 5123		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 61 2 11 3 11 4 22 3 13 4 12 4 1

Sample Output

141 21 32 33 4

這題我就不吐槽了。。範例都是錯的。。我沒看討論區。。一直不知道。。浪費我那麼多時間調試。。我去。。

各種吐血。。我也逗比。。範例明顯的出現環。。

正確的範例應該是：

1

3

1 2

1 3

3 4

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<vector>#include<queue>#include<cmath>using namespace std;const int maxn1 = 15050;//邊數的最大值const int maxn2 = 1050;//頂點個數的最大值int f[maxn2];//f[i]為頂點i在集合對樹中的根節點int s[maxn1];//記錄選擇的邊的序號int n, m;//集線器的個數， 邊的個數int cnt;//選擇的邊的數目int ans;//記錄最大的長度struct Edge{    int u;    int v;    int len;};Edge edge[maxn1];//邊的數組bool cmp(Edge a, Edge b)//按長度從小到大的排序{    return a.len<b.len;}void init()//初始化{    for(int i=0; i<=n; i++)        f[i] = i;}int find(int x)//並查集的find函數{    return f[x] == x? x:f[x]=find( f[x] );}void kruskal(){    int x, y;    cnt = 0;    for(int i=1; i<=m; i++)    {        x = find( edge[i].u );        y = find( edge[i].v );        if( x==y )   continue;        f[y] = x;        ans = edge[i].len;        cnt++;        s[cnt] = i;        if( cnt>=n-1 ) break;    }}void output()//輸出函數{    printf("%d\n", ans);    printf("%d\n", cnt);    for(int i=1; i<=cnt; i++)        printf("%d %d\n", edge[ s[i] ].u, edge[ s[i] ].v);}int main(){    while(scanf("%d%d", &n, &m)!=EOF)    {        init();        for(int i=1; i<=m; i++)            scanf("%d%d%d", &edge[i].u, &edge[i].v, &edge[i].len);        sort( edge+1, edge+m+1, cmp );        kruskal();        output();    }    return 0;}

POJ 1861：Network（最小產生樹&&kruskal）