poj3177——Redundant Paths//最小割边

跟3352一模一样，不过，需加判重。

#include<iostream>#include<cstdio>#include<cstring>using namespace std;int n,r;#define maxn 5005bool map[maxn][maxn];class node{public:int to,next;};node g[maxn*5];int head[maxn],cnt;int dfn[maxn],low[maxn],du[maxn];void add(int u,int v){g[++cnt].to =v;g[cnt].next =head[u];head[u]=cnt;}void dfs(int u,int fa){dfn[u]=low[u]=++cnt;int i;for(i=head[u];i;i=g[i].next ){int v=g[i].to;if(!dfn[v]){dfs(v,u);low[u]=min(low[u],low[v]);}else if(v!=fa){low[u]=min(low[u],dfn[v]);}}}void solve(){int i,j,ans=0;memset(dfn,0,sizeof(dfn));memset(low,0,sizeof(low));cnt=0;dfs(1,-1);memset(du,0,sizeof(du));for(i=1;i<=n;i++){for(j=head[i];j;j=g[j].next ){int v=g[j].to;if(low[i]!=low[v]) du[low[i]]++;//low值相同的节点即在一个双联通分量里}}for(i=1;i<=n;i++){//printf("%d ",du[i]);if(du[i]==1)ans++;}//printf("/n");//for(i=1;i<=n;i++)//printf("%d ",low[i]);//printf("/n");printf("%d/n",(ans+1)/2);}int main(){int a,b;while(scanf("%d%d",&n,&r)!=EOF){cnt=0;memset(map,0,sizeof(map));while(r--){scanf("%d%d",&a,&b);if(!map[a][b]){add(a,b); add(b,a);map[a][b]=map[b][a]=1;}}solve();}return 0;}