Luo Gu problem solution [P2341] [HAOI2006] popular cattle

Face questions

answer

\ (Tarjan \) after point reduction statistics of each point.

If there are a plurality of points to \ (0 \) , direct output \ (0 \) , or the degree of output \ (0 \) is the number of points in the strongly connected components.

Code

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cctype>
#define gI gi
#define itn int
#define File(x) freopen(x".in","r",stdin);freopen(x".out","w",stdout)

using namespace std;

inline int gi()
{
    int f = 1, x = 0; char c = getchar();
    while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
    while (c >= '0' && c <= '9') {x = x * 10 + c - '0'; c = getchar();}
    return f * x;
}

int n, m, tot, head[100003], nxt[100003], ver[100003], vis[100003];
int x, y, dfn[100003], low[100003], sta[100003], sy[100003], ys[100003];
int num, cnt, sum, ans, kok;
int cd[100003];

inline void add(int u, int v)
{
    ver[++tot] = v, nxt[tot] = head[u], head[u] = tot;
}

void Tarjan(int u)//Tarjan缩点
{
    dfn[u] = low[u] = ++num; sta[++cnt] = u; vis[u] = 1;
    for (int i = head[u]; i; i = nxt[i])
    {
        int v = ver[i];
        if (!dfn[v])
        {
            Tarjan(v);
            low[u] = min(low[u], low[v]);
        }
        else if (vis[v]) low[u] = min(low[u], dfn[v]);
    }
    if (dfn[u] == low[u])
    {
        int y;
        ++kok;
        do
        {
            y = sta[cnt--];
            vis[y] = 0;
            sy[y] = kok;
            ++ys[kok];//统计强连通分量里点的个数
        } while (y != u);
    }
}

int main()
{
    //File("P2341");
    n = gi(), m = gi();
    for (int i = 1; i <= m; i+=1)
    {
        int u = gi(), v = gi();
        add(u, v);
    }
    for (int i = 1; i <= n; i+=1) if (!dfn[i]) Tarjan(i);
    for (int i = 1; i <= n; i+=1)
    {
        for (int j = head[i]; j; j = nxt[j])
        {
            int u = ver[j];
            if (sy[i] != sy[u])
            {
                ++cd[sy[i]];//统计出度
            }
        }
    }
    int fl = 0;
    for (int i = 1; i <= kok; i+=1)
    {
        if (!cd[i])//出度为0
        {
            if (fl) {puts("0"); return 0;}//有多个出度为0的点,直接输出0
            fl = i;//记录答案所在的强连通分量的编号
        }
    }
    printf("%d\n", ys[fl]);//输出答案
    return 0;
}