洛谷 P1111 修复公路

题目传送门

解题思路:

本题说让任意两个点都可以互相到达,其实就是这n个点连接成一个联通块,一开始有n个联通块,每当有两个原本不在同一联通块的点连在一起,连通块数量减一,直到只剩一块.

AC代码:

#include<cstdio>
#include<iostream>
#include<algorithm> 

using namespace std;

int n,m,fa[1001],tot = 1;
bool p[1001];
struct kkk {
    int from,to,t;
}e[100001];

inline int find_father(int x) {
    if(fa[x] == x) return x;
    return fa[x] = find_father(fa[x]);
} 

bool cmp(kkk a,kkk b) {
    return a.t < b.t;
}

int main()
{
    scanf("%d%d",&n,&m);
    for(int i = 1;i <= n; i++) fa[i] = i;
    for(int i = 1;i <= m; i++) 
        scanf("%d%d%d",&e[i].from,&e[i].to,&e[i].t);
    sort(e+1,e+1+m,cmp);
    for(int i = 1;i <= m; i++) {
        int x = find_father(e[i].from);
        int y = find_father(e[i].to);
        if(x != y) {
            tot++;
            fa[x] = y;
        }
        if(tot == n) {
            printf("%d",e[i].t);
            return 0;
        }
    }
    printf("-1");
    return 0;
}