/*
ID: wuqi9395@126.com
PROG:
LANG: C++
*/
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<cmath>
#include<cstdio>
#include<vector>
#include<string>
#include<fstream>
#include<cstring>
#include<ctype.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define INF (1<<30)
#define PI acos(-1.0)
#define mem(a, b) memset(a, b, sizeof(a))
#define rep(i, a, n) for (int i = a; i < n; i++)
#define per(i, a, n) for (int i = n - 1; i >= a; i--)
#define eps 1e-6
#define debug puts("===============")
#define pb push_back
#define mkp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
#define POSIN(x,y) (0 <= (x) && (x) < n && 0 <= (y) && (y) < m)
typedef long long ll;
typedef unsigned long long ULL;
const int maxn = 1100;
vector<int> g[maxn];
int dfn[maxn], low[maxn], sccno[maxn], dfs_clock, scc_cnt;
stack<int> s;
void dfs(int u) {
dfn[u] = low[u] = ++ dfs_clock;
s.push(u);
for (int i = 0; i < g[u].size(); i++) {
int v = g[u][i];
if (!dfn[v]) {
dfs(v);
low[u] = min(low[u], low[v]); //通过儿子更新low值
}
else if (!sccno[v]) low[u] = min(low[u], dfn[v]); //通过回边更新low值(且回边访问的点必须不在已划分的强连通分量中)
}
if (low[u] == dfn[u]) { //得到强连通分量
scc_cnt++;
while(1) {
int x = s.top(); s.pop();
sccno[x] = scc_cnt;
if(x == u) break;
}
}
}
void find_scc(int n) {
dfs_clock = scc_cnt = 0;
memset(sccno, 0, sizeof(sccno));
memset(dfn, 0, sizeof(dfn));
while(!s.empty()) s.pop();
for (int i = 1; i <= n; i++) if (!dfn[i]) dfs(i);
}
const int maxm = 6100;
int mp[maxm][2];
vector<int> newg[maxn];
int degree[maxn];
void build_new_map(int m) {
for (int i = 1; i <= scc_cnt; i++) newg[i].clear(), degree[i] = 0;
for (int i = 0; i < m; i++) {
int u = sccno[mp[i][0]], v = sccno[mp[i][1]];
if (u != v) {
degree[v]++;
newg[u].pb(v);
}
}
}
bool toposort() {
queue<int> q;
for (int i = 1; i <= scc_cnt; i++) if (!degree[i]) q.push(i);
if (q.size() > 1) return 0;
int tot = 0;
while(!q.empty()) {
int u = q.front();
q.pop();
for (int i = 0; i < newg[u].size(); i++) {
int v = newg[u][i];
degree[v]--;
if (!degree[v]) q.push(v);
}
if (q.size() > 1) return 0;
}
return 1;
}
int main () {
int n, m;
int t;
scanf("%d", &t);
while(t--) {
scanf("%d%d", &n, &m);
for (int i = 0; i <= n; i++) g[i].clear();
int u, v;
rep(i, 0, m) {
scanf("%d%d", mp[i], mp[i] + 1);
g[mp[i][0]].pb(mp[i][1]);
}
find_scc(n);
build_new_map(m);
if (toposort()) puts("Yes");
else puts("No");
}
return 0;
}
POJ 2762 Going from u to v or from v to u? (有向图求单连通性)
原文:http://blog.csdn.net/sio__five/article/details/39033957