[Bzoj 1051] [Haoi 2006] Popular cattle (Tarjan)

http://www.lydsy.com:808/JudgeOnline/problem.php?id=1051

Well... This question seems to have been seen on POJ?

Compare the water problem, very good to come up with ideas. The relationship between cows and cows is like a graph, the cow a likes the Cow B is equivalent to the establishment of the A->b, and then in this direction graph, each strong connected component of the cows are like each other, the image is shrunk into dag,dag if there is only one point with a degree of 0, This point is a popular bull for all the cows in the Chingqiang connected component, if there is no 0 point, of course, there is no popular cattle, if the number of points of 0 is greater than 1, then each of the strong connected components out of 0 The number of fans will be less than n-1, there is no popular cattle.

A bit of a hole in the title, seemingly on the subject of the data range is small, was the pit WA once.

#include <iostream> #include <stdio.h> #include <string.h> #include <algorithm> #include < stdlib.h> #define Maxe 100050#define maxv 500050using namespace std;struct edge{int u,v,next;} Edges[maxe];int n,m;int head[maxv],ncount=0;int low[maxv],dfn[maxv],belong[maxv],stack[maxv],top=0;int cnt=0,tot=0 ; tot= number of strong connected components int NUM[MAXV],OUTDEGREE[MAXV];    Num[i]= the number of points in the strong connected component of I, outdegree= the degree of bool visit[maxv];void addedge (int u,int V) {edges[++ncount].u=u;    Edges[ncount].v=v;    Edges[ncount].next=head[u]; Head[u]=ncount;}    void Tarjan (int u)//tarjan indent {low[u]=dfn[u]=++cnt;    Stack[++top]=u;    Visit[u]=true;        for (int p=head[u];p!=-1;p=edges[p].next) {int v=edges[p].v;            if (!dfn[v]) {Tarjan (v);        Low[u]=min (Low[u],low[v]);    } else if (Visit[v]) low[u]=min (Low[u],dfn[v]);        } if (Dfn[u]==low[u]) {tot++;        int v=-1;         while (U!=V) {v=stack[top--];   Belong[v]=tot;            num[tot]++;        Visit[v]=false;    }}}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);    Addedge (U,V);    } for (int i=1;i<=n;i++) if (!dfn[i]) Tarjan (i);    for (int i=1;i<=m;i++) if (belong[edges[i].u]!=belong[edges[i].v]) outdegree[belong[edges[i].u]]++; int Res=0,ans;            Res= the number of points with a degree of 0, ans= a strong connected component number of 0 for (int i=1;i<=tot;i++) if (outdegree[i]==0) {res++;        Ans=i;    } if (res==1) printf ("%d\n", Num[ans]);    else printf ("0\n"); return 0;}



[Bzoj 1051] [Haoi 2006] Popular cattle (Tarjan)