POJ 2762 Going from u to v or from v to u?- Tarjan

Description

判断一个有向图是否对于任意两点 $x$,  $y$ 都有一条路径使$x - >y$或 $y - >x$

Solution

对于一个强联通分量内的点 都是可以互相到达的。

接下来我们考虑缩点后的DAG是否任意两点都有路径能使一点到达另一点。

然后我就不会了~~

我们进行一遍拓扑排序， 如果过程中有超过一个点的入度为 $0$ ，那么就不符合条件（仔细想想好像还是对的

Code

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<queue>
 #define rd read()
 #define R register
 using namespace std;

 const int N = 1e5;

 int head[N], tot;
 int Head[N], Tot;
 int dfn[N], low[N], st[N] ,tp, vis[N], cnt;
 int col_num, col[N], ru[N];
 int n, m, T;

 queue<int> q;

 struct edge {
     int nxt, to, fr;
 }e[N << ], E[N << ];

 int read() {
     int X = , p = ; char c = getchar();
     for(; c > '' || c < ''; c = getchar()) if(c == '-') p = -;
     for(; c >= '' && c <= ''; c = getchar()) X = X *  + c - '';
     return X * p;
 }

 void add(int u, int v) {
     e[++tot].to = v;
     e[tot].nxt = head[u];
     e[tot].fr = u;
     head[u] = tot;
 }

 void Add(int u, int v) {
     E[++Tot].to = v;
     E[Tot].nxt = Head[u];
     E[Tot].fr = u;
     Head[u] = Tot;
 }

 bool topsort() {
     int num = ;
     for(int i = ; i <= tot; ++i) {
         int u = col[e[i].fr], v = col[e[i].to];
         if(u == v) continue;
         ru[v]++;
         Add(u, v);
     }
     for(int i = ; i <= col_num; ++i)
         if(ru[i] == ) num++, q.push(i);
     if(num > )
         return ;
     for(int u; !q.empty();) {
         if(num > ) return ;
         u = q.front(); q.pop();
         num--;
         for(int i = Head[u]; i; i = E[i].nxt) {
             int nt = E[i].to;
             ru[nt]--;
             if(!ru[nt])
                 num++, q.push(nt);
         }
     }
     return ;
 }

 void tarjan(int u) {
     dfn[u] = low[u] = ++cnt;
     st[++tp] = u;
     vis[u] = ;
     for(R int i = head[u]; i; i = e[i].nxt) {
         int nt = e[i].to;
         if(!dfn[nt]) {
             tarjan(nt);
             low[u] = min(low[u], low[nt]);
         }
         else if(vis[nt]) low[u] = min(low[u], dfn[nt]);
     }
     if(low[u] == dfn[u]) {
         ++col_num;
         for(; tp; ) {
             int z = st[tp--];
             vis[z] = ;
             col[z] = col_num;
             if(z == u) break;
         }
     }
 }

 void init() {
     Tot = tp = tot = cnt = col_num = ;
     memset(vis,  ,sizeof(vis));
     memset(dfn, , sizeof(dfn));
     memset(col, , sizeof(col));
     memset(head, , sizeof(head));
     memset(low, , sizeof(low));
     memset(st, , sizeof(tp));
     memset(Head, , sizeof(Head));
     memset(ru, , sizeof(ru));
     while(!q.empty()) q.pop();
 }

 int main()
 {
     T = rd;
     for(; T; T--) {
         init();
         n = rd; m = rd;
         for(int i = ;  i <= m; ++i) {
             int u = rd, v = rd;
             add(u, v);
         }
         for(int i = ; i <= n; ++i)
             if(!dfn[i]) tarjan(i);
         if(topsort()) puts("Yes");
         else puts("No");
     }
 }