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 = , 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 = ; 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 << ]; struct node{
int u, v, c;
}Node[maxn << ]; 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, );
} bool bfs()
{
queue<int> Q;
mem(d, );
d[s] = ;
Q.push(s);
while(!Q.empty())
{
int u = Q.front(); Q.pop();
for(int i = head[u]; i!= -; i = nex[i])
{
int v = Node[i].v;
if(!d[v] && Node[i].c > )
{
d[v] = d[u] + ;
Q.push(v);
if(v == t) break;
}
}
}
return d[t] != ;
} int dfs(int u, int cap)
{
int ret = ;
if(u == t || cap == )
return cap;
for(int &i = cur[u];i != -; i = nex[i])
{
int v = Node[i].v;
if(d[v] == d[u] + && Node[i].c > )
{
int V = dfs(v, min(Node[i].c, cap));
Node[i].c -= V;
Node[i ^ ].c += V;
ret += V;
cap -= V;
if(cap == ) break;
}
}
return ret;
} int Dinic()
{
int ret = ;
while(bfs())
{
memcpy(cur, head, sizeof head);
ret += dfs(s, INF);
}
return ret;
} int graph[][]; int main()
{
int T;
rd(T);
while(T--)
{
int u, v;
mem(head, -);
cnt = ;
rd(n), rd(m);
mem(sccno, );
mem(pre, );
dfs_clock = scc_cnt = ;
for(int i = ; i <= n; i++) G[i].clear();
for(int i = ; i <= m; i++)
{
int u, v;
rd(u), rd(v);
G[u].push_back(v);
}
for(int i = ; i <= n; i ++) if(!pre[i]) dfs(i);
s = , t = maxn - ;
rap(u, , n)
{
for(int i = ; 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], );
}
}
rap(i, , scc_cnt)
add(s, i, ), add(scc_cnt + i, t, );
if(scc_cnt - Dinic() == )
printf("Yes\n");
else
printf("No\n"); } return ;
}