Milk Factory//排位2

2023-10-28 03:58

文章标签 factory 排位 milk

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Milk Factory//暴力?

题目

The milk business is booming! Farmer John’s milk processing factory consists of N processing stations, conveniently numbered 1…N (1≤N≤100), and N−1 walkways, each connecting some pair of stations. (Walkways are expensive, so Farmer John has elected to use the minimum number of walkways so that one can eventually reach any station starting from any other station). To try and improve efficiency, Farmer John installs a conveyor belt in each of its walkways. Unfortunately, he realizes too late that each conveyor belt only moves one way, so now travel along each walkway is only possible in a single direction! Now, it is no longer the case that one can travel from any station to any other station.

However, Farmer John thinks that all may not be lost, so long as there is at least one station i such that one can eventually travel to station i from every other station. Note that traveling to station i from another arbitrary station j may involve traveling through intermediate stations between i and j. Please help Farmer John figure out if such a station i exists.

Input
The first line contains an integer N, the number of processing stations. Each of the next N−1 lines contains two space-separated integers ai and bi with 1≤ai,bi≤N and ai≠bi. This indicates that there is a conveyor belt that moves from station ai to station bi, allowing travel only in the direction from ai to bi.

Output
If there exists a station i such that one can walk to station i from any other station, then output the minimal such i. Otherwise, output −1.

Example
inputCopy
3
1 2
3 2
outputCopy
2
题意
给你连接两地的单向路径，求是否存在一个点可以到达所有其它点，输出最小的那个，没有则为-1

思路

利用了并查集里寻根的思想，能走到的地方+1，最后看哪个地方为n-1

代码

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <deque>
#include <queue>
#include <stack>
#define pi 3.1415926
using namespace std;
typedef long long ll;
const ll mod = 1e9+7;
int root[200];
int num[200];
int main(){int n;cin>>n;memset(num,0,sizeof(num));for(int i=1;i<=n;i++){root[i]=i;}int a,b;for(int i=1;i<=n-1;i++){cin>>a>>b;root[a]=b;}for(int i=1;i<=n;i++){int temp=i;while(root[temp]!=temp&&temp!=0){num[root[temp]]++;temp=root[temp];}}int ans=-1;for(int i=1;i<=n;i++)if(num[i]==n-1){ans=i;break;}printf("%d\n",ans);return 0;
}