UVA - 11709 Trust groups （强连通分量）

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;}