2377:Bad Cowtractors(最大生成树）

2377:Bad Cowtractors

总时间限制: 1000ms 内存限制: 65536kB
描述
Bessie has been hired to build a cheap internet network among Farmer John’s N (2 <= N <= 1,000) barns that are conveniently numbered 1..N. FJ has already done some surveying, and found M (1 <= M <= 20,000) possible connection routes between pairs of barns. Each possible connection route has an associated cost C (1 <= C <= 100,000). Farmer John wants to spend the least amount on connecting the network; he doesn’t even want to pay Bessie.

Realizing Farmer John will not pay her, Bessie decides to do the worst job possible. She must decide on a set of connections to install so that (i) the total cost of these connections is as large as possible, (ii) all the barns are connected together (so that it is possible to reach any barn from any other barn via a path of installed connections), and (iii) so that there are no cycles among the connections (which Farmer John would easily be able to detect). Conditions (ii) and (iii) ensure that the final set of connections will look like a “tree”.
输入
* Line 1: Two space-separated integers: N and M

• Lines 2..M+1: Each line contains three space-separated integers A, B, and C that describe a connection route between barns A and B of cost C.
  输出
• Line 1: A single integer, containing the price of the most expensive tree connecting all the barns. If it is not possible to connect all the barns, output -1.
  样例输入
  5 8
  1 2 3
  1 3 7
  2 3 10
  2 4 4
  2 5 8
  3 4 6
  3 5 2
  4 5 17
  样例输出
  42
  提示
  OUTPUT DETAILS:

The most expensive tree has cost 17 + 8 + 10 + 7 = 42. It uses the following connections: 4 to 5, 2 to 5, 2 to 3, and 1 to 3.

分析：

原来最大生成树就是并查集的问题，虽然不是很理解，但是还是要记住用法。后面的判断是t==n-1，而不是t==n，或者也可以一个个的判断父亲是否都是一个。

#include<iostream>#include<algorithm>using namespace std;//http://bailian.openjudge.cn/practice/2377/struct node{    int a,b,c;};int n,m,par[2000],t,sum;//t用于判断是否所有节点都被树包含，sum是最大生成树的和 node path[50000];bool cmp(node x,node y){    return x.c>y.c;}int find(int x){    if(x==par[x])return x;    else return find(par[x]);}void merge(int x,int y,int z){    int r1=find(x),r2=find(y);    if(r1==r2)return;    else{        par[r1]=r2;        sum+=z;        t++;    }}int main(){    cin>>n>>m;    t=0;sum=0;    for(int i=1;i<=n;i++){        par[i]=i;//并查集的初始化     }    for(int i=1;i<=m;i++){        cin>>path[i].a>>path[i].b>>path[i].c;    }     sort(path+1,path+m+1,cmp);    for(int i=1;i<=m;i++){        merge(path[i].a,path[i].b,path[i].c);    }    if(t==n-1){        cout<<sum<<endl;    }    else cout<<"-1"<<endl;}