hdu4612 Warm up[边双连通分量缩点+树的直径]

给你一个连通图，你可以任意加一条边，最小化桥的数目。

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添加一条边，发现在边双内是不会减少桥的。只有在边双与边双之间加边才有效。于是，跑一遍边双并缩点，然后就变成一棵树，这样要加一条非树边，路径上的点（边双）就都变成一个大边双了，所以问题转化为求树上最长路径，跑直径即可。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 #include<queue>
 #define mst(x) memset(x,0,sizeof x)
 #define dbg(x) cerr << #x << " = " << x <<endl
 #define dbg2(x,y) cerr<< #x <<" = "<< x <<"  "<< #y <<" = "<< y <<endl
 using namespace std;
 typedef long long ll;
 typedef double db;
 typedef pair<int,int> pii;
 template<typename T>inline T _min(T A,T B){return A<B?A:B;}
 template<typename T>inline T _max(T A,T B){return A>B?A:B;}
 template<typename T>inline char MIN(T&A,T B){return A>B?(A=B,):;}
 template<typename T>inline char MAX(T&A,T B){return A<B?(A=B,):;}
 template<typename T>inline void _swap(T&A,T&B){A^=B^=A^=B;}
 template<typename T>inline T read(T&x){
     x=;int f=;char c;while(!isdigit(c=getchar()))if(c=='-')f=;
     while(isdigit(c))x=x*+(c&),c=getchar();return f?x=-x:x;
 }
 const int N=2e5+,M=1e6+;
 struct thxorz{int to,nxt;}G1[M<<],G2[N<<];
 int Head1[N],Head2[N],tot1,tot2;
 int n,m;
 inline void Addedge(int x,int y){
     G1[++tot1].to=y,G1[tot1].nxt=Head1[x],Head1[x]=tot1;
     G1[++tot1].to=x,G1[tot1].nxt=Head1[y],Head1[y]=tot1;
 }
 inline void Addtedge(int x,int y){
     G2[++tot2].to=y,G2[tot2].nxt=Head2[x],Head2[x]=tot2;
     G2[++tot2].to=x,G2[tot2].nxt=Head2[y],Head2[y]=tot2;
 }
 #define y G1[j].to
 int dfn[N],low[N],cut[M<<],cnt;
 void tarjan(int x,int i){
     dfn[x]=low[x]=++cnt;
     for(register int j=Head1[x];j;j=G1[j].nxt)if(j^(i^)){
         if(!dfn[y]){
             tarjan(y,j);MIN(low[x],low[y]);
             if(low[y]>dfn[x])cut[j]=cut[j^]=;
         }
         else MIN(low[x],dfn[y]);
     }
 }
 int bel[N],dcc;
 void dfs(int x){
     bel[x]=dcc;
     for(register int j=Head1[x];j;j=G1[j].nxt)if(!bel[y]&&!cut[j])dfs(y);
 }
 #undef y
 #define y G2[j].to
 int dep,pt;
 void tree_dfs(int x,int fa,int d){
     if(MAX(dep,d))pt=x;
     for(register int j=Head2[x];j;j=G2[j].nxt)if(y^fa)tree_dfs(y,x,d+);
 }
 #undef y
 int main(){//freopen("test.in","r",stdin);//freopen("test.ans","w",stdout);
     while(read(n),read(m),n||m){
         tot1=;dcc=tot2=cnt=dcc=;mst(Head1),mst(Head2),mst(dfn),mst(low),mst(cut),mst(bel);
         for(register int i=,x,y;i<=m;++i)read(x),read(y),Addedge(x,y);
         for(register int i=;i<=n;++i)if(!dfn[i])tarjan(i,);
         for(register int i=;i<=n;++i)if(!bel[i])++dcc,dfs(i);
         for(register int t=,u,v;t<=tot1;t+=){
             u=G1[t].to,v=G1[t^].to;
             if(bel[u]^bel[v])Addtedge(bel[u],bel[v]);//dbg2(bel[u],bel[v]);
         }
         dep=,tree_dfs(,,);dep=,tree_dfs(pt,,);
         printf("%d\n",dcc-dep);
     }
     return ;
 }

总结：和桥有关的可以首先考虑缩点建树，因为树有很好的性质，并且树边都是有桥连接起来的，同一个边双方便处理。