UVA 796 Critical Links（模板题）(无向图求桥)

<题目链接>

题目大意：

无向连通图求桥，并将桥按顺序输出。

解题分析；

无向图求桥的模板题，下面用了kuangbin的模板。

 #include <cstdio>
 #include <iostream>
 #include <cstring>
 #include <algorithm>
 #include <map>
 #include <vector>
 using namespace std;
 const int N = 1e4+;
 const int M = 1e5+;
 struct Edge{
     int to,next;
     bool cut;//是否为桥的标记
 }edge[M];
 int head[N],tot,n;
 int low[N],dfn[N],Stack[N];
 int Index,top;
 bool Instack[N],cut[N];
 int add_block[N];//删除一个点后增加的连通块
 int bridge;
 void init(){
     memset(head,-,sizeof(head));
     memset(dfn,,sizeof(dfn));
     memset(Instack,false,sizeof(Instack));
     memset(add_block,,sizeof(add_block));
     memset(cut,false,sizeof(cut));
     Index=top=bridge=tot = ;
 }
 void addedge(int u,int v){
     edge[tot].to = v;edge[tot].next = head[u];edge[tot].cut = false;
     head[u] = tot++;
 }
 void Tarjan(int u,int pre){
     low[u] = dfn[u] = ++Index;
     Stack[top++] = u;
     Instack[u] = true;
     int son = ;
     for(int i = head[u];i != -;i = edge[i].next){
         int v = edge[i].to;
         if(v == pre)continue;
         if( !dfn[v] ){
             son++;
             Tarjan(v,u);
             if(low[u] > low[v])low[u] = low[v];
             if(low[v] > dfn[u]){    //一条无向边(u,v)是桥，当且仅当(u,v)为树枝边，且满足DFS(u)<low(v)
                 bridge++;
                 edge[i].cut = true;     //正反两边都标记为桥
                 edge[i^].cut = true;
             }
             if(u != pre && low[v] >= dfn[u]){
                 cut[u] = true;   //该点为割点
                 add_block[u]++;
             }
         }
         else if( low[u] > dfn[v])low[u] = dfn[v];
     }
     if(u == pre && son > )cut[u] = true;  //若u为根，且分支数>1，则u割点
     if(u == pre)add_block[u] = son - ;
     Instack[u] = false;
     top--;
 }
 void solve(){
     for(int i = ;i <= n;i++)
         if( !dfn[i] )
             Tarjan(i,i);
     printf("%d critical links\n",bridge);

     vector<pair<int,int> >ans;          /*--   将桥按顺序输出  --*/
     for(int u = ;u <= n;u++)
         for(int i = head[u];i != -;i = edge[i].next)
             if(edge[i].cut && edge[i].to > u){
                 ans.push_back(make_pair(u,edge[i].to));
             }
     sort(ans.begin(),ans.end());
     for(int i = ;i < ans.size();i++)
         printf("%d - %d\n",ans[i].first-,ans[i].second-);
     printf("\n");
 }
 //处理重边
 /*map<int,int>mapit;
 inline bool isHash(int u,int v)
 {
     if(mapit[u*N+v])return true;
     if(mapit[v*N+u])return true;
     mapit[u*N+v] = mapit[v*N+u] = 1;
     return false;
 }*/
 int main(){
     while(scanf("%d",&n) == ){
         init();
         int u,k,v;
         //mapit.clear();
         for(int i = ;i <= n;i++){
             scanf("%d (%d)",&u,&k);
             u++;
             //这样加边，要保证正边和反边是相邻的，建无向图
             while(k--){
                 scanf("%d",&v);
                 v++;
                 if(v <= u)continue;
                 //if(isHash(u,v))continue;
                 addedge(u,v);
                 addedge(v,u);
             }
         }
         solve();
     }
     return ;
 }

2018-10-18