虽然本题是求单连通的，但是我们需要先求强连通分量，因为，强连通分量中存在双向路径，因此可以缩点，缩点后就好处理多了。
如果要满足题意，缩点后的树必须是一条链，而且所有边的方向都是一样的，如果出现分支，很容易证明会出现不可到达的一对点。
那么剩下的就是求最长链的顶点数是否等于强连通分量的个数了。
那么就可以使用拓扑排序，或者直接DFS。
拓扑排序中，不能出现同时有两个点的入度为0。
DFS从入度为0的顶点开始搜，是求出深搜的层数，因为单链的话层数是等于顶点数的
下面贴出用Kosaraju算法实现的代码
/*
ID: sdj22251
PROG: subset
LANG: C++
*/
#include <iostream>
#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 <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cmath>
#include <ctime>
#define LOCA
#define MAXN 10005
#define INF 100000000
#define eps 1e-7
using namespace std;
struct Edge
{int v, next;
}edge[10 * MAXN], revedge[10 * MAXN];
int head[MAXN], revhead[MAXN], e, visited[MAXN];
int order[MAXN], cnt, id[MAXN], tmp;
int uu[5 * MAXN], vv[5 * MAXN], in[MAXN];
int n, m;
void init()
{e = 0;memset(head, -1, sizeof(head));memset(revhead, -1, sizeof(revhead));memset(in, 0 , sizeof(in));
}
void insert(const int &x, const int &y)
{edge[e].v = y;edge[e].next = head[x];head[x] = e;revedge[e].v = x;revedge[e].next = revhead[y];revhead[y] = e;e++;
}
void readdata()
{for(int i = 0; i < m; i++){scanf("%d%d", &uu[i], &vv[i]);insert(uu[i], vv[i]);}
}
void dfs(int u) //搜索原图
{visited[u] = 1;for(int i = head[u]; i != -1; i = edge[i].next){int v = edge[i].v;if(!visited[v])dfs(v);}order[cnt++] = u;
}
void dfs_rev(int u) //搜索逆图
{visited[u] = 1;id[u] = cnt;for(int i = revhead[u]; i != -1; i = revedge[i].next){int v = revedge[i].v;if(!visited[v])dfs_rev(v);}
}
void search(int u, int deep) //缩点后的搜索
{visited[u] = 1;tmp = max(tmp, deep);for(int i = head[u]; i != -1; i = edge[i].next){int v = edge[i].v;if(!visited[v]){if(id[u] == id[v])search(v, deep);else search(v, deep + 1);}}
}
void Kosaraju()
{init();readdata();memset(visited, 0, sizeof(visited));cnt = 0;for(int i = 1; i <= n; i++){if(!visited[i])dfs(i);}memset(visited, 0, sizeof(visited));cnt = 0;for(int i = n - 1; i >= 0; i--){if(!visited[order[i]]){cnt++;dfs_rev(order[i]);}}for(int i = 0; i < m; i++){int u = id[uu[i]];int v = id[vv[i]];if(u != v) in[v]++;}tmp = 0;memset(visited, 0, sizeof(visited));for(int i = 1; i <= n; i++){if(in[id[i]] == 0){search(i, 1);break;}}
}
int main()
{int T;scanf("%d", &T);while(T--){scanf("%d%d", &n, &m);Kosaraju();int ans = 0;if(tmp != cnt) printf("No\n");else printf("Yes\n");}return 0;
}
tarjan代码
#include <iostream>
#include <map>
#include <cstdio>
#include <stack>
#include <cstring>
#include <algorithm>
#define MAXN 10005
#define MAXM 100005
#define INF 1000000000
using namespace std;
int n, m;
int scc;//强连通分量
int index;//每个节点的dfs访问次序编号
int dfn[MAXN];//标记结点i的dfs访问次序
int low[MAXN];//记录节点u或u的子树中的所有节点的最小标号
int fa[MAXN];//属于哪个分支
bool instack[MAXN];//是否在栈中
int in[MAXN], head[MAXN], e;
int out[MAXN];//出度
stack <int>s;
int tmp;
int vis[MAXN];
struct Edge
{int v, next;
}edge[MAXM];
void insert(int x, int y)
{edge[e].v = y;edge[e].next = head[x];head[x] = e++;
}
void tarjan(int u)
{dfn[u] = low[u] = ++index;s.push(u);instack[u] = true;for (int j = head[u]; j != -1; j = edge[j].next){int v = edge[j].v;if(dfn[v] == 0)//未曾访问过{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++;while(1){int tmp = s.top();s.pop();instack[tmp] = 0;fa[tmp] = scc;if(tmp == u) break;}}
}
void init()
{scc = index = 0;memset(dfn, 0, sizeof(dfn));memset(instack, 0, sizeof(instack));e = 0;memset(head, -1, sizeof(head));memset(in, 0, sizeof(in));memset(out, 0, sizeof(out));}
void dfs(int v, int deep)
{tmp = max(tmp, deep);vis[v] = 1;for(int i = head[v]; i != -1; i = edge[i].next)if(!vis[edge[i].v]){if(fa[edge[i].v] == fa[v]) dfs(edge[i].v, deep);else dfs(edge[i].v, deep + 1);}
}
void solve()
{for (int i = 1;i <= n; i++){if (!dfn[i])tarjan(i);}for (int i = 1;i <= n; i++){for(int j = head[i]; j != -1; j = edge[j].next){int u = fa[i];int v = fa[edge[j].v];if(u != v){out[u]++;in[v]++;}}}tmp = 0;memset(vis, 0, sizeof(vis));for(int i = 1; i <= n; i++)if(in[fa[i]] == 0){dfs(i, 1);break;}printf(tmp == scc ? "Yes\n" : "No\n");
}int main()
{int T, x, y;scanf("%d", &T);while(T--){init();scanf("%d%d", &n, &m);while(m--){scanf("%d%d", &x, &y);insert(x, y);}solve();}return 0;
}