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 #include 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; }
