强连通分量分解 tarjan算法 (hdu 1269)

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强连通分量分解 tarjan算法 (hdu 1269)

题意：
给出一个有n个点m条边的有向图，判断该图是否只有一个强连通分量。

限制：
0 <= N <= 10000
0 <= M <= 100000

思路：
tarjan算法分解强连通分量。

/*强连通分量分解 tarjan算法 (hdu 1269)  题意：  给出一个有n个点m条边的有向图，判断该图是否只有一个强连通分量。  限制：  0 <= N <= 10000  0 <= M <= 100000 */#include<iostream>#include<cstdio>#include<stack>#include<vector>#include<cstring>using namespace std;const int MAX_V = 1e5+5;#define PB push_backvector<int> G[MAX_V];stack<int> tarjan_stack;int DFN[MAX_V], LOW[MAX_V];int tarjan_set[MAX_V];bool in_stack[MAX_V];int scc_cnt, tarjan_cnt;void tarjan(int u){DFN[u] = LOW[u] = ++tarjan_cnt;tarjan_stack.push(u);in_stack[u] = true;for(int i = 0; i < G[u].size(); ++i){int ch = G[u][i];if(!DFN[ch]){tarjan(ch);LOW[u] = min(LOW[u], LOW[ch]);} else if(in_stack[ch]){LOW[u] = min(LOW[u], DFN[ch]);}}if(DFN[u] == LOW[u]){int tmp;++scc_cnt;do{tmp = tarjan_stack.top(); tarjan_stack.pop();in_stack[tmp] = false;tarjan_set[tmp] = scc_cnt;}while(tmp != u);}}void add_edge(int u,int v){G[u].PB(v);}void init_tarjan(int n){scc_cnt = tarjan_cnt = 0;memset(DFN, 0, sizeof(DFN));memset(LOW, 0, sizeof(LOW));memset(in_stack, 0, sizeof(in_stack));memset(tarjan_set, 0, sizeof(tarjan_set));for(int i = 0; i <= n; ++i)G[i].clear();}//点的序号从0开始void gao(int n){for(int i = 0; i < n; ++i){if(!DFN[i]) tarjan(i);}for(int i = 1; i < n; ++i){if(tarjan_set[i] != tarjan_set[i-1]){puts("No");return ;}}puts("Yes");}int main(){int n, m;while(scanf("%d%d", &n, &m) && (n || m)){init_tarjan(n);for(int i = 0; i < m; ++i){int u, v;scanf("%d%d", &u, &v);add_edge(u-1, v-1);}gao(n);}return 0;}

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