codeforces 655D D. Robot Rapping Results Report(拓撲排序+拓撲序記錄)

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題目連結：

D. Robot Rapping Results Report

time limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard output

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.

Examplesinput

4 5
2 1
1 3
2 3
4 2
4 3

output

4

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.

題意：問給你一些數對,第一個數在第二個數的前邊,問是否能確定唯一的拓撲序,如果能問前多少數對就能確定,不能輸出-1;

思路：把圖和每個點的入度存好後就找入度為0的點,再把與這個點能到其他的點的入度-1;一直到所有點的入度都為0,可以用隊列存入度為0的點進行最佳化;如果隊列的元素多於1個就輸出-1;還有就是把兩個拓撲序相鄰的點的這條邊存起來,最後遍曆一遍輸入的資訊,恰好把這些邊減完就是最少的數對數量了;

AC代碼：

#include <bits/stdc++.h>using namespace std;const int N=1e5+4;int a[N],b[N],n,m,ind[N],mp[N];queue<int>qu;vector<int>v[N];int main(){    scanf("%d%d",&n,&m);    for(int i=1;i<=m;i++)    {        scanf("%d%d",&a[i],&b[i]);        //mp[a[i]][b[i]]=1;        ind[b[i]]++;        v[a[i]].push_back(b[i]);    }    int num=0,cnt=0;    for(int i=1;i<=n;i++)    {        if(!ind[i])        {            num++;            qu.push(i);        }    }    while(!qu.empty())    {        if(num>1)        {            cout<<"-1";            return 0;        }        int x=qu.front();        qu.pop();        num--;        int len=v[x].size();        for(int j=0;j<len;j++)        {            int t=v[x][j];            ind[t]--;            if(!ind[t])            {                qu.push(t);                num++;                mp[t]=x;                cnt++;            }        }    }    for(int i=1;i<=m;i++)    {        if(mp[b[i]]==a[i])           {               cnt--;               if(cnt==0)               {                   cout<<i<<"\n";                   break;               }           }    }    return 0;}

 

codeforces 655D D. Robot Rapping Results Report(拓撲排序+拓撲序記錄)