poj 2186 Popular Cows（Tarjan）

最新推荐文章于 2020-06-03 15:50:20 发布

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最新推荐文章于 2020-06-03 15:50:20 发布

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本文介绍了一种算法，用于解决牛群中寻找最受欢迎牛的问题。通过构建图模型并运用Tarjan算法进行强连通分量分析，最终确定能够被所有牛认为是流行的牛的数量。

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Popular Cows

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 24548		Accepted: 10072

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

Sample Output

Hint

Cow 3 is the only cow of high popularity.

Source

USACO 2003 Fall

给定牛之间的仰慕关系且仰慕是可传递的问能受到所有牛仰慕的牛的数目。。

如果要受到所有牛的仰慕则这个牛的点对于其他牛来说都是可达的

首先对图进行缩点得到DAG 如果在该DAG中只有一个叶子节点

则这个节点都可以由其他节点到达输出这个点所代表的联通分量中点的个数

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string.h>
#include <string>
#include <vector>
#include <queue>

#define MEM(a,x) memset(a,x,sizeof a)
#define eps 1e-8
#define MOD 10009
#define INF 99999999
#define ll __int64
#define bug cout<<"here"<<endl
#define fread freopen("ceshi.txt","r",stdin)
#define fwrite freopen("out.txt","w",stdout)

using namespace std;

void read(int &x)
{
    char ch;
    x=0;
    while(ch=getchar(),ch!=' '&&ch!='\n')
    {
        x=x*10+ch-'0';
    }
}

const int MAXN=10010;
const int MAXM=50010;
struct Edge
{
    int to,next;
}edge[MAXM];
int head[MAXN],tot;
int low[MAXN],dfn[MAXN],sta[MAXN],belong[MAXN];
int index,top;
int scc;
bool instack[MAXN];
int num[MAXN];
int n,m;

void addedge(int u,int v)
{
    edge[tot].to=v; edge[tot].next=head[u]; head[u]=tot++;
}

void Tarjan(int u)
{
//    bug;
    int v;
    low[u]=dfn[u]=++index;
    sta[top++]=u;
    instack[u]=1;
    for(int i=head[u];i!=-1;i=edge[i].next)
    {
//        bug;
        v=edge[i].to;
        if(!dfn[v])
        {
            Tarjan(v);
            low[u]=min(low[u],low[v]);
        }
        else if(instack[v])
            low[u]=min(low[u],dfn[v]);
    }
//    bug;
    if(low[u]==dfn[u])
    {
        scc++;
        do{
            v=sta[--top];
            instack[v]=0;
            belong[v]=scc;
            num[scc]++;
        }while(u!=v);
    }
}

int cd[MAXN];
void solve()
{
//    bug;
    MEM(dfn,0);
    MEM(instack,0);
    MEM(cd,0);
    MEM(num,0);
    index=scc=top=0;
    for(int i=1;i<=n;i++)
        if(!dfn[i])
            Tarjan(i);
    for(int u=1;u<=n;u++)
    {
        for(int i=head[u];i!=-1;i=edge[i].next)
        {
            int v=edge[i].to;
            if(belong[u]!=belong[v])
                cd[belong[u]]++;
        }
    }
    int cnt=0;
    int x;
    for(int i=1;i<=scc;i++)
    {
        if(cd[i]==0)
        {
            cnt++;
            x=i;
        }
    }
    if(cnt==1)
        printf("%d\n",num[x]);
    else puts("0");
}
void init()
{
    tot=0;
    MEM(head,-1);
}

int main()
{
//    fread;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        init();
        int u,v;
        for(int i=0;i<m;i++)
        {
            scanf("%d%d",&u,&v);
            addedge(u,v);
        }
        solve();
    }
    return 0;
}