poj2186 Popular Cows（强连通）

最新推荐文章于 2018-09-29 20:12:23 发布

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最新推荐文章于 2018-09-29 20:12:23 发布

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原文链接：http://www.cnblogs.com/wenruo/p/5007977.html

本文介绍了一个利用Tarjan算法解决牛群中崇拜传递性问题的方法。通过求解强连通分量，找到所有牛共同崇拜的对象。文章提供了完整的C++代码实现，并解释了如何判断解的存在性。

崇拜有传递性。求所有牛都崇拜的牛
tarjan算法求强连通。

如果不连通就不存在。如果联通，缩点后唯一一个出度为零的点就是答案，有多个则不存在。

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <complex>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <cassert>
using namespace std;

const int N = 10005;
const int M = 100005;

struct Edge {
    int to, next;
} edge[M];
int head[N];
int cnt_edge;

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

int dfn[N];
int low[N];
int stk[N];
bool in[N];
int kind[N];

int top;
int idx;
int cnt;

int n, m;

void dfs(int u)
{
    dfn[u] = low[u] = ++idx;
    in[u] = true;
    stk[++top] = u;
    for (int i = head[u]; i != -1; i = edge[i].next)
    {
        int v = edge[i].to;
        if (!dfn[v])
        {
            dfs(v);
            low[u] = min(low[v], low[u]);
        }
        else if(in[v])
        {
            low[u] = min(low[u], dfn[v]);
        }
    }

    if (low[u] == dfn[u])
    {
        ++cnt;
        int j;
        do {
            j = stk[top--];
            in[j] = false;
            kind[j] = cnt;
        } while (j != u);
    }
}

void init()
{
    memset(dfn, 0, sizeof dfn);
    memset(head, -1, sizeof head);
    cnt_edge = 0;
    top = cnt = idx = 0;
}


int deg[N];

void solve()
{
    // 缩点后出度为0的点个数为1
    for (int u = 1; u <= n; ++u)
    {
        int k = kind[u];
        for (int i = head[u]; i != -1; i = edge[i].next)
        {
            int v = edge[i].to;
            if (kind[u] != kind[v]) deg[k]++;
        }
    }

    int flag = 0;
    int p;
    for (int i = 1; i <= cnt; ++i)
    {
        if (deg[i] == 0)
        {
            flag++;
            p = i;
            if (flag > 1) break;
        }
    }
    if (flag == 1)
    {
        int ans = 0;
        for (int i = 1; i <= n; ++i) if (kind[i] == p) ++ans;
        printf("%d\n", ans);
    }
    else
    {
        printf("0\n");
    }
}

int main()
{
    while (~scanf("%d%d", &n, &m))
    {
        if (n == 0 && m == 0) break;
        int a, b;
        init();
        for (int i = 0; i < m; ++i)
        {
            scanf("%d%d", &a, &b);
            add_edge(a, b);
        }

        for (int i = 1; i <= n; ++i)
        {
            if (!dfn[i]) dfs(i);
        }

        solve();
    }
    return 0;
}

转载于:https://www.cnblogs.com/wenruo/p/5007977.html