标签:ack sizeof struct map cos -- exp print main
1 //无向图求割边 2 #include <cstdio> 3 #include <cstdlib> 4 #include <cstring> 5 #include <cmath> 6 #include <algorithm> 7 #include <queue> 8 #include <stack> 9 #include <vector> 10 #include <deque> 11 #include <map> 12 #include <iostream> 13 using namespace std; 14 typedef long long LL; 15 const double pi = acos(-1.0); 16 const double e = exp(1); 17 //const int MAXN =2e5+10; 18 const LL N = 1000000007; 19 20 struct edge{ 21 int w; 22 int to; 23 int next; 24 }edge[200009]; 25 int head[200009]; 26 27 int dfn[100009]; 28 int low[100009]; 29 int cnt = 1; 30 31 void tarjan(int u, int fa) 32 { 33 34 int i; 35 low[u] = dfn[u] = cnt++; 36 for(i = head[u]; i != -1; i = edge[i].next) 37 { 38 //cout << u << " " << fa << endl; 39 int v = edge[i].to; 40 if(!dfn[v]) 41 { 42 //cout << "** " << v << " " << u << endl; 43 tarjan(v, u); 44 low[u] = min(low[u], low[v]); 45 if(low[v] > dfn[u]) 46 { 47 printf("%d ---> %d\n", u, v); 48 } 49 } 50 else if(v != fa) 51 { 52 low[u] = min(low[u], dfn[v]); 53 } 54 } 55 } 56 57 int main() 58 { 59 int n,m, i; 60 int cnt1, a, b; 61 while(scanf("%d%d", &n,&m) != EOF) 62 { 63 if(n == 0 && m == 0) 64 break; 65 cnt1 = 0; 66 cnt = 1; 67 memset(head, -1, sizeof(head)); 68 memset(dfn, 0, sizeof(dfn)); 69 memset(low, 0, sizeof(low)); 70 while(m--) 71 { 72 scanf("%d%d", &a,&b); 73 74 edge[cnt1].to = b; 75 edge[cnt1].next = head[a]; 76 head[a] = cnt1++; 77 78 edge[cnt1].to = a; 79 edge[cnt1].next = head[b]; 80 head[b] = cnt1++; 81 } 82 for(i = 1; i <= n; i++) 83 { 84 if(!dfn[i]) 85 { 86 tarjan(i, 0); 87 } 88 } 89 90 } 91 return 0; 92 }
标签:ack sizeof struct map cos -- exp print main
原文地址:https://www.cnblogs.com/daybreaking/p/12778850.html