Going from u to v or from v to u? POJ - 2762（强连通 有向最长路径）

In order to make their sons brave, Jiajia and Wind take them to a big cave. The cave has n rooms, and one-way corridors connecting some rooms. Each time, Wind choose two rooms x and y, and ask one of their little sons go from one to the other. The son can either go from x to y, or from y to x. Wind promised that her tasks are all possible, but she actually doesn't know how to decide if a task is possible. To make her life easier, Jiajia decided to choose a cave in which every pair of rooms is a possible task. Given a cave, can you tell Jiajia whether Wind can randomly choose two rooms without worrying about anything?

Input

The first line contains a single integer T, the number of test cases. And followed T cases. 

The first line for each case contains two integers n, m(0 < n < 1001,m < 6000), the number of rooms and corridors in the cave. The next m lines each contains two integers u and v, indicating that there is a corridor connecting room u and room v directly. 

Output

The output should contain T lines. Write 'Yes' if the cave has the property stated above, or 'No' otherwise.

Sample Input

1
3 3
1 2
2 3
3 1

Sample Output

Yes

问：
给出的图是否存在对任意的u 和 v 要么u -> v, 要么 v -> u，这两者是或者的关系 不是并且
先求强连通，因为每个强连通分量中的点都可以相互到达，然后缩点 求最小路径覆盖
网上都是用的拓扑求最长路径  但最小路径覆盖就是这个思想
所以最后只要判断是否只有一个最小路径即可
代码就是改了一下HDU 3861的输出   所以就是水题啦  

#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <map>
#include <cctype>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#include <bitset>
#define rap(i, a, n) for(int i=a; i<=n; i++)
#define rep(i, a, n) for(int i=a; i<n; i++)
#define lap(i, a, n) for(int i=n; i>=a; i--)
#define lep(i, a, n) for(int i=n; i>a; i--)
#define rd(a) scanf("%d", &a)
#define rlld(a) scanf("%lld", &a)
#define rc(a) scanf("%c", &a)
#define rs(a) scanf("%s", a)
#define rb(a) scanf("%lf", &a)
#define rf(a) scanf("%f", &a)
#define pd(a) printf("%d\n", a)
#define plld(a) printf("%lld\n", a)
#define pc(a) printf("%c\n", a)
#define ps(a) printf("%s\n", a)
#define MOD 2018
#define LL long long
#define ULL unsigned long long
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define _  ios_base::sync_with_stdio(0),cin.tie(0)
//freopen("1.txt", "r", stdin);
using namespace std;
const int maxn = 110000, INF = 0x7fffffff;

int n, m, s, t;

vector<int> G[maxn];
int pre[maxn], low[maxn], sccno[maxn], dfs_clock, scc_cnt;
stack<int> S;


void dfs(int u)
{
    pre[u] = low[u] = ++dfs_clock;
    S.push(u);
    for(int i = 0; i < G[u].size(); i ++)
    {
        int v = G[u][i];
        if(!pre[v])
        {
            dfs(v);
            low[u] = min(low[u], low[v]);
        }
        else if(!sccno[v])
        {
            low[u] = min(low[u], pre[v]);
        }
    }
    if(low[u] == pre[u])
    {
        scc_cnt++;
        for(;;)
        {
            int x = S.top(); S.pop();
            sccno[x] = scc_cnt;
            if(x == u) break;
        }
    }
}


int cur[maxn], head[maxn], cnt, d[maxn], nex[maxn << 1];

struct node{
    int u, v, c;
}Node[maxn << 1];

void add_(int u, int v, int c)
{
    Node[cnt].u = u;
    Node[cnt].v = v;
    Node[cnt].c = c;
    nex[cnt] = head[u];
    head[u] = cnt++;
}

void add(int u, int v, int c)
{
    add_(u, v, c);
    add_(v, u, 0);
}

bool bfs()
{
    queue<int> Q;
    mem(d, 0);
    d[s] = 1;
    Q.push(s);
    while(!Q.empty())
    {
        int u = Q.front(); Q.pop();
        for(int i = head[u]; i!= -1; i = nex[i])
        {
            int v = Node[i].v;
            if(!d[v] && Node[i].c > 0)
            {
                d[v] = d[u] + 1;
                Q.push(v);
                if(v == t) break;
            }
        }
    }
    return d[t] != 0;
}

int dfs(int u, int cap)
{
    int ret = 0;
    if(u == t || cap == 0)
        return cap;
    for(int &i = cur[u];i != -1; i = nex[i])
    {
        int v = Node[i].v;
        if(d[v] == d[u] + 1 && Node[i].c > 0)
        {
            int V = dfs(v, min(Node[i].c, cap));
            Node[i].c -= V;
            Node[i ^ 1].c += V;
            ret += V;
            cap -= V;
            if(cap == 0) break;
        }
    }
    return ret;
}

int Dinic()
{
    int ret = 0;
    while(bfs())
    {
        memcpy(cur, head, sizeof head);
        ret += dfs(s, INF);
    }
    return ret;
}

int graph[5010][5010];

int main()
{
    int T;
    rd(T);
    while(T--)
    {
        int u, v;
        mem(head, -1);
        cnt = 0;
        rd(n), rd(m);
        mem(sccno, 0);
        mem(pre, 0);
        dfs_clock = scc_cnt = 0;
        for(int i = 1; i <= n; i++) G[i].clear();
        for(int i = 1; i <= m; i++)
        {
            int u, v;
            rd(u), rd(v);
            G[u].push_back(v);
        }
        for(int i = 1; i <= n; i ++) if(!pre[i]) dfs(i);
        s = 0, t = maxn - 1;
        rap(u, 1, n)
        {
            for(int i = 0; i < G[u].size(); i ++)
            {
                int  v = G[u][i];
                 //cout << sccno[u] << "   " << sccno[v] << endl;
                if(sccno[u] != sccno[v])
                    add(sccno[u], scc_cnt + sccno[v], 1);
            }
        }
        rap(i, 1, scc_cnt)
            add(s, i, 1), add(scc_cnt + i, t, 1);
        if(scc_cnt - Dinic() == 1)
            printf("Yes\n");
        else
            printf("No\n");
           
    }

    return 0;
}