Description
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".
Input
* Line : Two space-separated integers: N and M * Lines ..M+: Each line contains three space-separated integers A, B, and C that describe a connection route between barns A and B of cost C.
Output
* Line : 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 -.
Sample Input
Sample Output
Hint
OUTPUT DETAILS: The most expensive tree has cost + + + = . It uses the following connections: to , to , to , and to .
Source
其实不过就是裸的最小生成树，（kruscal实现），只不过这里的话要将边按从大到小排序，然后判断一下是不是连通的就可以了。
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<math.h>
#include<algorithm>
#include<queue>
#include<set>
#include<bitset>
#include<map>
#include<vector>
#include<stdlib.h>
#include <stack>
using namespace std;
#define PI acos(-1.0)
#define max(a,b) (a) > (b) ? (a) : (b)
#define min(a,b) (a) < (b) ? (a) : (b)
#define ll long long
#define eps 1e-10
#define MOD 1000000007
#define N 1006
#define M 20006
#define inf 1e12
struct Node{
int x,y;
int cost;
}edge[M];
int n,m;
int fa[N];
void init(){
for(int i=;i<N;i++){
fa[i]=i;
}
}
int find(int x){
return fa[x]==x?x:fa[x]=find(fa[x]);
}
bool cmp(Node a,Node b){
return a.cost>b.cost;
}
int main()
{
while(scanf("%d%d",&n,&m)==){
init();
for(int i=;i<m;i++){
int a,b,c;
scanf("%d%d%d",&edge[i].x,&edge[i].y,&edge[i].cost);
}
sort(edge,edge+m,cmp);
int ans=;
int num=n-;
for(int i=;i<m;i++){
int root1=find(edge[i].x);
int root2=find(edge[i].y);
if(root1!=root2){
ans+=edge[i].cost;
fa[root1]=root2;
num--;
}
}
if(num!=){
printf("-1\n");
}
else{
printf("%d\n",ans);
}
}
return ;
}