cf-645D. Robot Rapping Results Report(拓扑序列)

645D. Robot Rapping Results Report

题目点击打开链接

While Farmer John rebuilds his farm in an unfamiliar portion of Bovinia, Bessie is out trying some alternative jobs. In her new gig as a reporter, Bessie needs to know about programming competition results as quickly as possible. When she covers the 2016 Robot Rap Battle Tournament, she notices that all of the robots operate under deterministic algorithms. In particular, robot i will beat robot j if and only if robot i has a higher skill level than robot j. And if robot i beats robot j and robot j beats robot k, then robot i will beat robot k. Since rapping is such a subtle art, two robots can never have the same skill level.

Given the results of the rap battles in the order in which they were played, determine the minimum number of first rap battles that needed to take place before Bessie could order all of the robots by skill level.

Input

The first line of the input consists of two integers, the number of robots n (2 ≤ n ≤ 100 000) and the number of rap battles m ().

The next m lines describe the results of the rap battles in the order they took place. Each consists of two integers ui and vi (1 ≤ ui, vi ≤ n,ui ≠ vi), indicating that robot ui beat robot vi in the i-th rap battle. No two rap battles involve the same pair of robots.

It is guaranteed that at least one ordering of the robots satisfies all m relations.

Output

Print the minimum k such that the ordering of the robots by skill level is uniquely defined by the first k rap battles. If there exists more than one ordering that satisfies all m relations, output -1.

Examples

input

output

input

3 2
1 2
3 2

output

-1

Note

In the first sample, the robots from strongest to weakest must be (4, 2, 1, 3), which Bessie can deduce after knowing the results of the first four rap battles.

In the second sample, both (1, 3, 2) and (3, 1, 2) are possible orderings of the robots from strongest to weakest after both rap battles.

【题意】：

给出一个有向图，确定最小的k，通过前k条边就能确定唯一的拓扑序列。若整个图不存在唯一的拓扑序列，输出-1

【分析】

跑一遍拓扑排序，判断是否有唯一拓扑序列，同时记录这个拓扑序列用到的编号最大的边

//#include<bits/stdc++.h>
#include <algorithm>
#include <cstring>
#include <iostream>
#include <cstdio>
#include <string>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include<cmath>
#define maxn 100005
using namespace std;
using namespace std;
const int N=202020;
struct node{int u,v,next;
}e[N];
int head[N],cnt;
int n,m;
int q[N];
int in[N];
//void add(int u,int v)
//{
//  cout<<"cnt="<<cnt<<endl;
//    //e[cnt]=node{u,v,head[u]};
//    e[cnt].u=u;
//  e[cnt].v=v;
//  e[cnt].next=head[u]; cout<<"e[cnt].u="<<e[cnt].u<<"e[cnt].v="<<e[cnt].v<<"e[cnt].next="<<e[cnt].next<<endl;
//    head[u]=cnt++;cout<<"head[u]="<<head[u]<<endl;
//}
int main()
{while(cin>>n>>m){memset(head,-1,sizeof(head));cnt=0;memset(in,0,sizeof(in));for(int i=0;i<m;i++){int u,v;scanf("%d%d",&u,&v);
//            add(u,v);e[cnt].u=u;e[cnt].v=v;e[cnt].next=head[u]; head[u]=cnt++;in[v]++;}int ans=0;int tail=0,flag=1;for(int i=1;i<=n;i++)if(in[i]==0)q[tail++]=i;if(tail>1)flag=0;for(int i=0;i<tail;i++){int t=q[i];int res=0;for(int j=head[t];~j;j=e[j].next)//注意~j{//cout<<"j="<<j<<endl;if(--in[e[j].v]==0){q[tail++]=e[j].v;//cout<<"e[j].v="<<e[j].v<<endl;res++;ans=max(ans,j); //记录编号最大边}//    cout<<"e[j].next="<<e[j].next<<"res="<<res<<endl;}if(res>1)flag=0;}if(flag==0||tail<n)cout<<-1<<endl;  //不唯一else  {cout<<ans+1<<endl;    //拓扑唯一}
}
return 0;
}

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cf-645D. Robot Rapping Results Report(拓扑序列)

645D. Robot Rapping Results Report

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