POJ3352Road Construction(无向图强连通)
http://poj.org/problem?id=3352
无向图强连通分量缩点 知道一个等式:
若要使得任意一棵树,在增加若干条边后,变成一个双连通图,那么
至少增加的边数 =( 这棵树总度数为1的结点数 + 1 )/ 2
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<stack> 5 #include<algorithm> 6 #include<stdlib.h> 7 using namespace std; 8 #define N 1010 9 #define M 2010 10 struct node 11 { 12 int u,v,next; 13 }edge[M]; 14 stack<int>s; 15 int head[N],pre[N],low[N],sccno[N],scc,dep,t,de[N],vis[N],w[N][N]; 16 void init() 17 { 18 t = 0; 19 memset(head,-1,sizeof(head)); 20 scc=0;dep=0; 21 memset(sccno,0,sizeof(sccno)); 22 memset(low,0,sizeof(low)); 23 memset(pre,0,sizeof(pre)); 24 memset(vis,0,sizeof(vis)); 25 } 26 void add(int u,int v) 27 { 28 edge[t].u = u; 29 edge[t].v = v; 30 edge[t].next = head[u]; 31 head[u] = t++; 32 } 33 void dfs(int u,int fa) 34 { 35 low[u] = pre[u] = ++dep; 36 s.push(u); 37 for(int i = head[u] ; i!=-1 ; i = edge[i].next) 38 { 39 int v = edge[i].v; 40 if(v==fa) continue; 41 if(!pre[v]) 42 { 43 dfs(v,u); 44 low[u] = min(low[u],low[v]); 45 } 46 else if(!sccno[v]) 47 low[u] = min(low[u],pre[v]); 48 } 49 if(low[u]==pre[u]) 50 { 51 scc++; 52 for(;;) 53 { 54 int x = s.top();s.pop(); 55 sccno[x] = scc; 56 if(x==u) break; 57 } 58 } 59 } 60 int main() 61 { 62 int i,n,m,kk=0; 63 char str[100]; 64 while(cin>>n>>m) 65 { 66 kk++;init(); 67 memset(w,0,sizeof(w)); 68 memset(de,0,sizeof(de)); 69 int a,b; 70 while(m--) 71 { 72 scanf("%d%d",&a,&b); 73 add(a,b);add(b,a); 74 } 75 dfs(1,1); 76 for(i = 0 ; i < t ; i++) 77 { 78 int u = edge[i].u,v = edge[i].v; 79 if(!w[u][v]&&sccno[u]!=sccno[v]) 80 { 81 w[u][v] = w[v][u]= 1; 82 de[sccno[u]]++; 83 de[sccno[v]]++; 84 } 85 } 86 int num=0; 87 for(i = 1; i <= scc ; i++) 88 if(de[i]==1) num++; 89 cout<<(num+1)/2<<endl; 90 } 91 return 0; 92 }