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F. The Meeting Place Cannot Be Changed

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Petr is a detective in Braginsk. Somebody stole a huge amount of money from a bank and Petr is to catch him. Somebody told Petr that some luxurious car moves along the roads without stopping.

Petr knows that it is the robbers who drive the car. The roads in Braginsk are one-directional and each of them connects two intersections. Petr wants to select one intersection such that if the robbers continue to drive the roads indefinitely, they will sooner or later come to that intersection. The initial position of the robbers is unknown. Find such an intersection that fits the requirements.

Input

The first line of the input contains two integers?nn?and?mm?(2≤n≤1052≤n≤105,?2≤m≤5?1052≤m≤5?105)?— the number of intersections and the number of directed roads in Braginsk, respectively.

Each of the next?mm?lines contains two integers?uiui?and?vivi?(1≤ui,vi≤n1≤ui,vi≤n,?ui≠viui≠vi)?— the start and finish of the?ii-th directed road. It is guaranteed that the robbers can move along the roads indefinitely.

Output

Print a single integer?kk?— the intersection Petr needs to choose. If there are multiple answers, print any. If there are no such intersections, print??1?1.

Examples

input

Copy

5 6 1 2 2 3 3 1 3 4 4 5 5 3

output

Copy

3

input

Copy

3 3 1 2 2 3 3 1

output

Copy

1

Note

?

In the first example the robbers can move, for example, along the following routes:?(1?2?3?1)(1?2?3?1),?(3?4?5?3)(3?4?5?3),?(1?2?3?4?5?3?1)(1?2?3?4?5?3?1). We can show that if Petr chooses the?33-rd intersection, he will eventually meet the robbers independently of their route.

題意：
劫匪的初始位置未知，他沿著道路不停的移動，城市的道路是單向的，并且每條道路都連接兩個交叉路口，
選擇一個交叉路口，如果劫匪繼續無限期地開車，他們遲早會到達那個交叉路口。

分析：

其實這個問題可以看做是一個強連通分量里面的所有簡單環是否存在至少一個公共點
那么我們假設一個點滿足條件就等價于從這個點出發，一定不會滯留在一個死循環中，
而是最后回到了出發點.對所有路徑都滿足。如果這個點不滿足，那么應該再去找哪個點呢？

這個點滯留在死循環中，那么就說明它不屬于這個死循環的環，所以接下來的那個點一定是這兩個環的交點，

因為該點絕對滿足不會進入這兩個環的死循環，接下來就是這樣找下去，直到遇到交點是已經遇到過的，

那么說明有兩個交集點為0的環，則輸出-1.由于這樣找在有解的情況是最快的，

所以時間一多說明幾乎是無解的,則可以限時查詢的時間。

#include<bits/stdc++.h> using namespace std; const int N=5e5+50;//數據量 int n,m;//節點數、邊數 int head[N],to[N],nxt[N],tot;//存儲有向圖 inline void add(int x,int y)//x為起點，y為終點建邊 {to[tot]=y;nxt[tot]=head[x];head[x]=tot++; } int vis[N];//標記已經訪問過的節點 int in[N],id[N]; bool dt[N],k; const bool cmp(int a,int b)//間接排序函數 {return in[a]<in[b];//按入度排序 } //計算搜索時間 inline double FIND_TIME() {return clock()/(double)CLOCKS_PER_SEC;} void dfs(int x) {vis[x]=1;//標記節點已經訪問過for(int i=head[x];~i;i=nxt[i]){if(!vis[to[i]])//下一個節點沒有訪問過dfs(to[i]);if(vis[to[i]]==1)//下一個節點訪問過k=1;if(k) return ;}vis[x]=2;//DFS回溯 } inline bool check(int x)//檢查每一個點 {memset(vis,0,sizeof(vis));vis[x]=2;k=0;for(int i=1;i<=n;i++)if(!vis[i]){dfs(i);if(k){for(int i=1;i<=n;i++)if(vis[i]!=1)dt[i]=1;return false;}}return true; }int main() {memset(head,-1,sizeof(head));scanf("%d%d",&n,&m);for(int i=1;i<=m;i++){int u,v;scanf("%d%d",&u,&v);add(u,v);//建邊++in[u];//更新度}for(int i=1;i<=n;i++)id[i]=i;//節點編號sort(id+1,id+1+n,cmp);//排序for(int i=1;i<=n;i++){if(FIND_TIME()>0.9) break;//時間限制為1sif(!dt[id[i]]&&check(id[i]))//此點合法{printf("%d\n",id[i]);//輸出路口的編號return 0;} }puts("-1");//不存在這樣的點return 0; }

用鄰接表來存儲有向圖

#include<bits/stdc++.h> using namespace std; const int mx = 1e5 + 10; int n,m,in[mx],ty,type[mx],d,mark; vector <int> vec[mx]; int dfn[mx],id[mx],size,is,sta[mx]; int vis[mx],flag; bool vic[mx]; void tarjan(int x) {dfn[x] = id[x] = ++is;vis[x] = 1;sta[++size] = x;for(int i=0;i<vec[x].size();i++){int son = vec[x][i];if(!dfn[son]){tarjan(son);id[x] = min(id[x],id[son]);}else if(vis[son]) id[x] = min(id[x],dfn[son]);}if(dfn[x]==id[x]){ty++;if(sta[size]!=x) d++,mark = ty;while(sta[size]!=x){type[sta[size]] = ty;vis[sta[size]] = 0;size--;}type[x] = ty,size--,vis[x] = 0;} } void dfs(int x,int num) {if(flag) return ;for(int i=0;i<vec[x].size();i++){if(flag) return;int son = vec[x][i];if(vis[son]==num){if(vic[son]) flag = -1;else flag = son,vic[son] = 1;}if(vis[son]<num) vis[son] = num,dfs(son,num);}vis[x] = num + 1; } double TIME(){//獲得單位運行時間return 1.0*clock()/CLOCKS_PER_SEC; } int check(int x,int num) {while(TIME()<0.5){vis[x] = num + 1;dfs(x,num);if(!flag) return x;if(flag==-1) return -1;x = flag,num += 2,flag = 0;}return -1; } int main() {scanf("%d%d",&n,&m);int a,b,c = 0,ans;for(int i=1;i<=m;i++){scanf("%d%d",&a,&b);vec[a].push_back(b);}for(int i=1;i<=n;i++)if(!dfn[i]) tarjan(i);if(d>1) puts("-1");else{for(int i=1;i<=n;i++)if(type[i]==mark){printf("%d\n",check(i,1));break;}}return 0; }

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