（連通圖 縮點 強聯通分支）Popular Cows -- poj --2186

標籤：

http://poj.org/problem?id=2186

 

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 31 22 12 3

Sample Output

1

 

先縮點（把強聯通分量看為一個點）， 判斷出度為 0 的點有幾個，如果大於 1 則輸出 0， 否則輸出 出度為零的點的個數

 

代碼：

#include<cstdio>#include<cstring>#include<cstdlib>#include<iostream>#include<queue>#include<stack>#include<algorithm>using namespace std;#define N 50005struct node{    int v, next;}a[N];int Head[N], cnt;int dfn[N], low[N], Time, bnt, belong[N];int Stack[N], InStack[N], top;void Init(){    cnt = Time = bnt = top = 0;    memset(Head, -1, sizeof(Head));    memset(dfn, 0, sizeof(dfn));    memset(low, 0, sizeof(low));    memset(Stack, 0, sizeof(Stack));    memset(InStack, 0, sizeof(InStack));}void Add(int u, int v){    a[cnt].v = v;    a[cnt].next = Head[u];    Head[u] = cnt++;}void Tarjar(int u){    int v;    low[u] = dfn[u] = ++Time;    InStack[u] = 1;    Stack[top++] = u;    for(int j=Head[u]; j!=-1; j=a[j].next)    {        v = a[j].v;        if(!dfn[v])        {            Tarjar(v);            low[u] = min(low[u], low[v]);        }        else if(InStack[v])            low[u] = min(low[u], dfn[v]);    }    if(dfn[u]==low[u])    {        bnt++;        do        {            v = Stack[--top];            InStack[v] = 0;            belong[v] = bnt;        }while(u!=v);    }}int main(){    int n, m;    while(scanf("%d%d", &n, &m)!=EOF)    {        int i, u, v;        Init();        for(i=1; i<=m; i++)        {            scanf("%d%d", &u, &v);            Add(u, v);        }        for(i=1; i<=n; i++)        {            if(!dfn[i])            Tarjar(i);        }        int Out[N]={0};        for(int i=1; i<=n; i++)        {            for(int j=Head[i]; j!=-1; j=a[j].next)            {                u = belong[i], v = belong[a[j].v];                if(u!=v)                    Out[u]++;            }        }        int flag=0, Index;        for(i=1; i<=bnt; i++)        {            if(!Out[i])            {                flag++;                Index = i;            }        }        if(flag>1)            printf("0\n");        else        {            int ans = 0;            for(i=1; i<=n; i++)            {                if(belong[i]==Index)                    ans++;            }            printf("%d\n", ans);        }    }    return 0;}

 

（連通圖 縮點 強聯通分支）Popular Cows -- poj --2186