P1111 修复公路
题目背景
A地区在地震过后,连接所有村庄的公路都造成了损坏而无法通车。
政府派人修复这些公路。
题目描述
给出A地区的村庄数N,和公路数M,公路是双向的。
并告诉你每条公路的连着哪两个村庄,并告诉你什么时候能修完这条公路。
问最早什么时候任意两个村庄能够通车,即最早什么时候任意两条村庄都存在至少一条修复完成的道路(可以由多条公路连成一条道路)
输入格式
第1行两个正整数N,M
下面M行,每行3个正整数x, y, t,告诉你这条公路连着x,y两个村庄,在时间t时能修复完成这条公路。
输出格式
如果全部公路修复完毕仍然存在两个村庄无法通车,则输出−1,否则输出最早什么时候任意两个村庄能够通车。
输入输出样例
输入 #1
4 4
1 2 6
1 3 4
1 4 5
4 2 3
输出 #1
5
说明/提示
N≤1000,M≤100000
x≤N,y≤N,t≤100000
Code
/*
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*/
// train —— 1111.cpp created by VB_KoKing on 2019-08-15.
/* Procedural objectives:
Variables required by the program:
Procedural thinking:
Functions required by the program:
Determination algorithm:
Determining data structure:
*/
/* My dear Max said:
"I like you,
So the first bunch of sunshine I saw in the morning is you,
The first gentle breeze that passed through my ear is you,
The first star I see is also you.
The world I see is all your shadow."
FIGHTING FOR OUR FUTURE!!!
*/
#include <cstdio>
#include <algorithm>
using namespace std;
struct Road{
int x, y, t;
}roads[100007];
int villages[100007];
int find(int x){
int res = x;
while (villages[res])
res = villages[res];
while (x != res){
int temp = villages[x];
villages[x] = res;
x = temp;
}
return res;
}
bool join(int x, int y){
int res_x = find(x), res_y = find(y);
if (res_x != res_y){
villages[res_x] = res_y;
return true;
}
return false;
}
bool cmp(const Road &a, const Road &b){
return a.t < b.t;
}
void solve() {
int n, m;
scanf("%d %d",&n, &m);
for (int i = 0; i < m; i++)
scanf("%d %d %d", &roads[i].x, &roads[i].y, &roads[i].t);
sort(roads, roads + m, cmp);
for (int i = 0; i < m; i++) {
n -= join(roads[i].x, roads[i].y);
if (n == 1){
printf("%d\n",roads[i].t);
return;
}
}
if (n > 1) printf("-1\n");
}
int main(){
solve();
return 0;
}