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,1):0;}
template<typename T>inline char MAX(T&A,T B){return A<B?(A=B,1):0;}
template<typename T>inline void _swap(T&A,T&B){A^=B^=A^=B;}
template<typename T>inline T read(T&x){
    x=0;int f=0;char c;while(!isdigit(c=getchar()))if(c=='-')f=1;
    while(isdigit(c))x=x*10+(c&15),c=getchar();return f?x=-x:x;
}
const int N=2e5+7,M=1e6+7;
struct thxorz{int to,nxt;}G1[M<<1],G2[N<<1];
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<<1],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^1)){
        if(!dfn[y]){
            tarjan(y,j);MIN(low[x],low[y]);
            if(low[y]>dfn[x])cut[j]=cut[j^1]=1;
        }
        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+1);
}
#undef y
int main(){//freopen("test.in","r",stdin);//freopen("test.ans","w",stdout);
    while(read(n),read(m),n||m){
        tot1=1;dcc=tot2=cnt=dcc=0;mst(Head1),mst(Head2),mst(dfn),mst(low),mst(cut),mst(bel);
        for(register int i=1,x,y;i<=m;++i)read(x),read(y),Addedge(x,y);
        for(register int i=1;i<=n;++i)if(!dfn[i])tarjan(i,0);
        for(register int i=1;i<=n;++i)if(!bel[i])++dcc,dfs(i);
        for(register int t=2,u,v;t<=tot1;t+=2){
            u=G1[t].to,v=G1[t^1].to;
            if(bel[u]^bel[v])Addtedge(bel[u],bel[v]);//dbg2(bel[u],bel[v]);
        }
        dep=0,tree_dfs(1,0,1);dep=0,tree_dfs(pt,0,1);
        printf("%d\n",dcc-dep);
    }
    return 0;
}

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。