POJ 2186 Popular Cows（强连通）

Posted 2020-08-09 贱人方

                                                              Popular Cows

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 30999		Accepted: 12580

Description

Every cow‘s dream is to become the most popular cow in the herd. In a herd of N (1 <= N <= 10,000) cows, you are given up to M (1 <= M <= 50,000) ordered pairs of the form (A, B) that tell you that cow A thinks that cow B is popular. Since popularity is transitive, if A thinks B is popular and B thinks C is popular, then A will also think that C is
popular, even if this is not explicitly specified by an ordered pair in the input. Your task is to compute the number of cows that are considered popular by every other cow.

Input

* Line 1: Two space-separated integers, N and M

* Lines 2..1+M: Two space-separated numbers A and B, meaning that A thinks B is popular.

Output

* Line 1: A single integer that is the number of cows who are considered popular by every other cow.

Sample Input

3 3
1 2
2 1
2 3

Sample Output

1

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#define inf 0x3f3f3f3f
#define mod 10000
typedef long long ll;
using namespace std;
const int N=10005;
const int M=100005;
int n,m,cnt,tim,top,cut;
int head[N],dfn[N],low[N],stack1[N];
int num[N],du[N],vis[N];
struct man {
    int to,next;
} edg[M];
void init() {
    memset(dfn,0,sizeof(dfn));
    memset(low,0,sizeof(low));
    memset(head,-1,sizeof(head));
    memset(vis,0,sizeof(vis));
    memset(num,0,sizeof(num));
    memset(du,0,sizeof(du));
    cnt=0;
    tim=1;
    top=0;
    cut=0;
}
void add(int u,int v) {
    edg[cnt].to=v;
    edg[cnt].next=head[u];
    head[u]=cnt;
    cnt++;
}
void dfs(int u,int fa) {
    dfn[u]=tim;
    low[u]=tim++;
    vis[u]=1;
    stack1[top++]=u;
    for(int i=head[u]; i!=-1; i=edg[i].next) {
        int v=edg[i].to;
        if(!vis[v]) {
            dfs(v,u);
            low[u]=min(low[u],low[v]);
        } else low[u]=min(low[u],low[v]);
    }
    if(low[u]==dfn[u]) {
        cut++;
        while(top>0&&stack1[top]!=u) {
            top--;
            vis[stack1[top]]=2;
            num[stack1[top]]=cut;
        }
    }
}
int main() {
    int u,v;
    while(~scanf("%d%d",&n,&m)) {
        init();
        for(int i=0; i<m; i++) {
            scanf("%d%d",&u,&v);
            add(u,v);
        }
        for(int i=1; i<=n; i++) {
            if(!vis[i])dfs(i,0);
        }
        for(int i=1; i<=n; i++) {
            for(int j=head[i]; j!=-1; j=edg[j].next) {
                if(num[i]!=num[edg[j].to])du[num[i]]++;
            }
        }
        int sum=0,x;
        for(int i=1; i<=cut; i++) {
            if(!du[i])sum++,x=i;
        }
        if(sum==1) {
            sum=0;
            for(int i=1; i<=n; i++) {
                if(num[i]==x)sum++;
            }
            cout<<sum<<endl;
        } else puts("0");
    }
    return 0;
}

tarjan

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#define inf 0x3f3f3f3f
#define mod 10000
typedef long long ll;
using namespace std;
const int N=10005;
const int M=100005;
struct Node
{
    int to,next;
}edge1[N*5],edge2[N*5];
//edge1用来存原图G，edge2用来存GT，即逆图
//都是用邻接表来储存的图，
int head1[N];//原图的邻接表的头结点
int head2[N];//逆图的邻接表的头结点
int mark1[N],mark2[N];
int tot1,tot2;
int cnt1,cnt2,st[N],belong[N];
int num,setNum[N];
struct Edge
{
    int l,r;
}e[N*5];//储存边

void add(int a,int b)//添加边a->b，在原图和逆图都需要添加
{
    edge1[tot1].to=b;edge1[tot1].next=head1[a];head1[a]=tot1++;//邻接表存的原图
    edge2[tot2].to=a;edge2[tot2].next=head2[b];head2[b]=tot2++;//邻接表存的逆图
}
void DFS1(int x)//深度搜索原图
{
    mark1[x]=1;
    for(int i=head1[x];i!=-1;i=edge1[i].next)
          if(!mark1[edge1[i].to])  DFS1(edge1[i].to);
    st[cnt1++]=x;//st数组是按照完成时间从小到大排序的
}
void DFS2(int x)//深度搜索逆图
{
    mark2[x]=1;
    num++;
    belong[x]=cnt2;
    for(int i=head2[x];i!=-1;i=edge2[i].next)
       if(!mark2[edge2[i].to]) DFS2(edge2[i].to);
}
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        tot1=tot2=1;
        for(int i=1;i<=n;i++)//初始化
        {
            head1[i]=head2[i]=-1;
            mark1[i]=mark2[i]=0;
        }
        for(int i=1;i<=m;i++)
        {
            int w,v;
            scanf("%d%d",&w,&v);
            e[i].l=w;e[i].r=v;//储存边
            add(w,v);//建立邻接表
        }
        cnt1=cnt2=1;
        for(int i=1;i<=n;i++)
        {
            if(!mark1[i])DFS1(i);
        }
        for(int i=cnt1-1;i>=1;i--)
        {
            if(!mark2[st[i]])
            {
                num=0;
                DFS2(st[i]);
                setNum[cnt2++]=num;
            }
        }
        int de[N];//计算出度
        memset(de,0,sizeof(de));
        for(int i=1;i<=m;i++)//计算各个DAG图的出度
        {
            if(belong[e[i].l]!=belong[e[i].r])//原图的边不属于同一连通分支
                de[belong[e[i].l]]++;
        }

        //计算DAG出度为0的个数
        int cnt=0,res;
        for(int i=1;i<cnt2;i++)
            if(!de[i]){cnt++;res=i;}
        if(cnt>1)  printf("0\n");
        else  printf("%d\n",setNum[res]);
    }
    return 0;
}

kosaraju