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链接 ：http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=28292
求强连通分量的个数。
#pragma comment(linker, "/STACK:10240000,10240000")
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <vector>
#include <cmath>
#include <queue>
#include <stack>
#include <set>
#include <map>
#define mod 4294967296
#define MAX 0x3f3f3f3f
#define lson o<<1, l, m
#define rson o<<1|1, m+1, r
#define SZ(x) ((int)ans.size())
#define MAKE make_pair
#define INFL 0x3f3f3f3f3f3f3f3fLL
#define mem(a) memset(a, 0, sizeof(a))
const double pi = acos(-1.0);
const double eps = 1e-9;
const int N = 100005;
const int M = 20005;
typedef long long ll;
using namespace std;
int n, m;
vector <int> G[N];
int pre[N], low[N], scc[N], dfs_clock, scc_cnt;
stack <int> S;
int T;
void dfs(int u) {
pre[u] = low[u] = ++dfs_clock;
S.push(u);
for(int i = 0; i < G[u].size(); i++) {
int v = G[u][i];
if(pre[v] == 0) {
dfs(v);
low[u] = min(low[u], low[v]);
} else if(scc[v] == 0) {
low[u] = min(low[u], pre[v]);
}
}
if(low[u] == pre[u]) {
scc_cnt++;
for(;;) {
int x = S.top(); S.pop();
scc[x] = scc_cnt;
if(x == u) break;
}
}
}
void find_scc() {
dfs_clock = scc_cnt = 0;
mem(scc);
mem(pre);
for(int i = 0; i < n; i++) {
if(pre[i] == 0) dfs(i);
}
}
map <string, int> mp;
int main() {
//freopen("in.txt","r",stdin);
while(cin >> n >> m && n + m) {
for(int i = 0; i < n; i++) {
G[i].clear();
}
string s, t;
for(int i = 0; i < n; i++) {
cin >> s >> t;
mp[s+t] = i;
}
for(int i = 0; i < m; i++) {
cin >> s >> t;
int x = mp[s+t];
cin >> s >> t;
int y = mp[s+t];
G[x].push_back(y);
}
find_scc();
cout << scc_cnt << endl;
}
return 0;
}
UVA - 11709 Trust groups （强连通分量）
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原文地址：http://blog.csdn.net/u013923947/article/details/45960385
