POJ 2762 Going from u to v or from v to u?（Tarjan + 拓扑排序）

题意：

给定有向图，问是否满足任意两点 x, y 使得 x->y 或 y->x 存在一条路径。

思路：

1. tarjan 算法缩点，然后重新建立缩点后的有向图 G0;

2. 对 G0 拓扑排序，看其是否为拓扑有序的，如果是输出 Yes, 如果不是输出 No;

#include <iostream>
#include <stack>
#include <vector>
#include <algorithm>
using namespace std;

const int MAXN = 1010;
vector<int> G[MAXN];
stack<int> S;
vector<int> G0[MAXN];
int dfn[MAXN], low[MAXN], sccno[MAXN], sccnum, cflag;
int indeg[MAXN];

void tarjan(int u) {
dfn[u] = low[u] = ++cflag;
S.push(u);

for (int i = 0; i < G[u].size(); i++) {
int v = G[u][i];
if (!dfn[v]) {
tarjan(v);
low[u] = min(low[u], low[v]);
} else if (!sccno[v]) {
low[u] = min(low[u], dfn[v]);
}
}

if (dfn[u] == low[u]) {
sccnum += 1;
int v = -1;
while (v != u) {
v = S.top();
S.pop();
sccno[v] = sccnum;
}
}
}

void findscc(int n) {
for (int i = 0; i <= n; i++)
dfn[i] = low[i] = sccno[i] = 0;
sccnum = cflag = 0;
for (int i = 1; i <= n; i++)
if (!dfn[i]) tarjan(i);
}

bool topsort(int n) {
stack<int> s;

for (int i = 1; i <= n; i++) {
if (!indeg[i]) s.push(i);
if (s.size() == 2) return false;
}
while (!s.empty()) {
int u = s.top();
s.pop();
bool flag = false;
for (int i = 0; i < G0[u].size(); i++) {
int v = G0[u][i];
indeg[v] -= 1;
if (!indeg[v]) {
if (flag) return false;
s.push(v);
flag = true;
}
}
}
return true;
}

int main() {
int cases;
scanf("%d", &cases);
while (cases--) {
int n, m;
scanf("%d%d", &n, &m);
for (int i = 0; i <= n; i++)
G[i].clear();
while (m--) {
int u, v;
scanf("%d%d", &u, &v);
G[u].push_back(v);
}
findscc(n);
for (int i = 0; i <= sccnum; i++)
G0[i].clear(), indeg[i] = 0;
for (int u = 1; u <= n; u++) {
for (int i = 0; i < G[u].size(); i++) {
int v = G[u][i];
if (sccno[u] != sccno[v]) {
indeg[sccno[v]] += 1;
G0[sccno[u]].push_back(sccno[v]);
}
}
}
if (topsort(sccnum))
printf("Yes\n");
else
printf("No\n");
}
return 0;
}