POJ 2762 Going from u to v or from v to u?_problem d: going from u to v or from v to u(poj276-优快云博客

本文链接：https://blog.youkuaiyun.com/K_ona/article/details/80963982

本文介绍了一种解决POJ 2762问题的方法，通过使用Tarjan算法进行缩点，并判断缩点后的图中是否存在出度或入度为0的强连通分量来解决问题。代码实现包括了初始化、Tarjan算法、构建缩点后的图以及最终的判断逻辑。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

缩点后判断出度或者入度为0的点的个数

#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<stack>

using namespace std;

const int maxn = 1006;

int n, m, inde, scc_num, res;
vector<int> P[maxn], G[maxn];
int low[maxn], dfn[maxn], ID[maxn], instack[maxn];
int in[maxn], out[maxn];
int M[maxn][maxn];
int dp[maxn];
stack<int> S;

void init()
{
    memset(dp, 0, sizeof(dp));
    memset(in, 0, sizeof(in));
    memset(out, 0, sizeof(out));
    memset(ID, 0, sizeof(ID));
    memset(low, -1, sizeof(low));
    memset(dfn, -1, sizeof(dfn));
    for(int i = 0; i < maxn; i++)
        P[i].clear(), G[i].clear();
    scc_num = inde = 0;
    while(!S.empty()) S.pop();
    res = 0;
}

void Get_DAG()
{
    for(int u = 1; u <= n; u++)
    {
        for(int i = 0; i < P[u].size(); i++)
        {
            int v = P[u][i];
            if(ID[u] != ID[v])
            {
                in[ID[v]]++;
                out[ID[u]]++;
                G[ID[u]].push_back(ID[v]);
            }
        }
    }
}

void tarjan(int u)
{
    low[u] = dfn[u] = ++inde;
    S.push(u);
    instack[u] = 1;
    for(int i = 0; i < P[u].size(); i++)
    {
        int v = P[u][i];
        if(dfn[v] == -1)
        {
            tarjan(v);
            low[u] = min(low[u], low[v]);
        }
        else if(instack[v])
            low[u] = min(low[u], dfn[v]);
    }

    if(dfn[u] == low[u])
    {
        ++scc_num;
        while(S.top() != u)
        {
            ID[S.top()] = scc_num;
            instack[S.top()] = 0;
            S.pop();
        }
        ID[u] = scc_num;
        instack[u] = 0;
        S.pop();
    }
}

int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);

    int T;
    cin >> T;
    while(T--)
    {
       init();
       cin >> n >> m;
       int u, v;
       for(int i = 0; i < m; i++)
       {
           cin >> u >> v;
           P[u].push_back(v);
       }

       for(int i = 1; i <= n; i++)
       {
           if(dfn[i] == -1)
                tarjan(i);
       }
       Get_DAG();
       for(int i = 1; i <= scc_num; i++)
        {
            if(!in[i])
                res++;
            if(!out[i])
                res++;
        }

        if(res <= 2)
            cout << "Yes" << endl;
        else
            cout << "No" << endl;
    }
    return 0;
}