Wormhole Sort//kruskal

题目

Farmer John’s cows have grown tired of his daily request that they sort themselves before leaving the barn each morning. They have just completed their PhDs in quantum physics, and are ready to speed things up a bit.

This morning, as usual, Farmer John’s N cows (1≤N≤105), conveniently numbered 1…N, are scattered throughout the barn at N distinct locations, also numbered 1…N, such that cow i is at location pi. But this morning there are also M wormholes (1≤M≤105), numbered 1…M, where wormhole i bidirectionally connects location ai with location bi, and has a width wi (1≤ai,bi≤N,ai≠bi,1≤wi≤109).

At any point in time, two cows located at opposite ends of a wormhole may choose to simultaneously swap places through the wormhole. The cows must perform such swaps until cow i is at location i for 1≤i≤N.

The cows are not eager to get squished by the wormholes. Help them maximize the width of the least wide wormhole which they must use to sort themselves. It is guaranteed that it is possible for the cows to sort themselves.

Input
The first line contains two integers, N and M.

The second line contains the N integers p1,p2,…,pN. It is guaranteed that p is a permutation of 1…N.
For each i between 1 and M, line i+2 contains the integers ai, bi, and wi.

Output
A single integer: the maximum minimal wormhole width which a cow must squish itself into during the sorting process. If the cows do not need any wormholes to sort themselves, output −1.

Examples
inputCopy
4 4
3 2 1 4
1 2 9
1 3 7
2 3 10
2 4 3
outputCopy
9
inputCopy
4 1
1 2 3 4
4 2 13
outputCopy
-1
Note
The first example is one possible way to sort the cows using only wormholes of width at least 9:

Cow 1 and cow 2 swap positions using the third wormhole. Cow 1 and cow 3 swap positions using the first wormhole. Cow 2 and cow 3 swap positions using the third wormhole.

The second example is no wormholes are needed to sort the cows.
题意
给两点和边权，各个牛回位置，要求最大边权最小值

思路

典型kruskal

代码

#include <cstdio>
#include <cstring>
#include <string>
#include <iostream>
#include <algorithm>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#define pi 3.1415926
using namespace std;
typedef long long ll;
const ll inf=100000000000;
int n,m,pos[100005],tot = 0,root[100005], ans = -1;
struct Edge{
    int from,to,dis;
}e[300005];
int findx(int x){
    if(root[x]==x) return x;
    else return root[x]=findx(root[x]);
}
void Union(int x, int y){
    root[findx(y)] = findx(x);
}
void add(int f,int t,int d){
    e[++tot].from = f;
    e[tot].to = t;
    e[tot].dis = d;
}
bool cmp(Edge a, Edge b){
    return a.dis>b.dis;
}
int main(){
    cin>>n>>m;
    for (int i=1;i<=n;i++)
    cin>>pos[i];
    for (int i=0;i<m;i++){
	    int a,b,c;
	    cin>>a>>b>>c;
	    add(a,b,c);
	    }
	    sort(e+1,e+tot+1,cmp);
	    int cnt = 1;
	    for (int i=1;i<=n;i++)
	    root[i] = i;
	    for (int i=1;i<=tot;i++){
	    while(findx(cnt)==findx(pos[cnt])) {
	        cnt++;
	        if (cnt>n) goto A;
	    }
	    if (findx(pos[e[i].from])!=findx(pos[e[i].to])){
	        Union(pos[e[i].from],pos[e[i].to]);
	        ans = e[i].dis;
	    }
    }
    A:printf("%d",ans);
    return 0;
}

注意