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 << 2], E[N << 2];

int read() {
    int X = 0, p = 1; char c = getchar();
    for(; c > '9' || c < '0'; c = getchar()) if(c == '-') p = -1;
    for(; c >= '0' && c <= '9'; c = getchar()) X = X * 10 + c - '0';
    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 = 0;
    for(int i = 1; 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 = 1; i <= col_num; ++i) 
        if(ru[i] == 0) num++, q.push(i);
    if(num > 1) 
        return 0;
    for(int u; !q.empty();) {
        if(num > 1) return 0;
        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 1;
}

void tarjan(int u) {
    dfn[u] = low[u] = ++cnt;
    st[++tp] = u;
    vis[u] = 1;
    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] = 0;
            col[z] = col_num;
            if(z == u) break;
        }
    }
}

void init() {
    Tot = tp = tot = cnt = col_num = 0;
    memset(vis, 0 ,sizeof(vis));
    memset(dfn, 0, sizeof(dfn));
    memset(col, 0, sizeof(col));
    memset(head, 0, sizeof(head));
    memset(low, 0, sizeof(low));
    memset(st, 0, sizeof(tp));
    memset(Head, 0, sizeof(Head));
    memset(ru, 0, sizeof(ru));
    while(!q.empty()) q.pop();
}

int main()
{
    T = rd;
    for(; T; T--) {
        init();
        n = rd; m = rd;
        for(int i = 1;  i <= m; ++i) {
            int u = rd, v = rd;
            add(u, v);
        }
        for(int i = 1; i <= n; ++i) 
            if(!dfn[i]) tarjan(i);
        if(topsort()) puts("Yes");
        else puts("No");
    }
}