Warm up
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 9073 Accepted Submission(s): 2120
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4612
Description:
N planets are connected by M bidirectional channels that allow instant transportation. It‘s always possible to travel between any two planets through these channels.
If we can isolate some planets from others by breaking only one channel , the channel is called a bridge of the transportation system.
People don‘t like to be isolated. So they ask what‘s the minimal number of bridges they can have if they decide to build a new channel.
Note that there could be more than one channel between two planets.
Input:
The input contains multiple cases.
Each case starts with two positive integers N and M , indicating the number of planets and the number of channels.
(2<=N<=200000, 1<=M<=1000000)
Next M lines each contains two positive integers A and B, indicating a channel between planet A and B in the system. Planets are numbered by 1..N.
A line with two integers ‘0‘ terminates the input.
Output:
For each case, output the minimal number of bridges after building a new channel in a line.
Sample Input:
4 4 1 2 1 3 1 4 2 3 0 0
Sample Output:
0
题意：
给出一个无向图，之后会加进来一条边，问加进来一条边过后，桥的最少数量为多少。
题解：
还是利用并查集先缩点，将无向图变为一颗树，树上每一条边都为桥。
然后加边我们希望尽量形成最大的环，那么考虑树的直径两端的点就满足了。
代码如下：
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <queue>
using namespace std;
typedef long long ll;
const int N = 2e5+5,M = 1e6+5;
int n,m;
int head[N];
struct Edge{
int u,v,next;
bool operator < (const Edge &A){
if(u==A.u) return v<A.v;
return u<A.u;
}
}e[M<<1],g[M<<1];
int T,tot,cnt;
int dfn[N],low[N],cut[N],num[N],f[N];
void adde(int u,int v){
e[tot].u=u;e[tot].v=v;e[tot].next=head[u];head[u]=tot++;
}
void init(){
T=0;tot=0;cnt=0;
memset(head,-1,sizeof(head));
memset(cut,0,sizeof(cut));
memset(dfn,0,sizeof(dfn));
memset(num,0,sizeof(num));
for(int i=0;i<=n+1;i++) f[i]=i;
}
int find(int x){
return f[x]==x?f[x]:f[x]=find(f[x]);
}
void Union(int x,int y){
int fx=find(x),fy=find(y);
if(fx!=fy) f[fx]=fy;
}
void Tarjan(int u,int pre){
dfn[u]=low[u]=++T;
int k=0;
for(int i=head[u];i!=-1;i=e[i].next){
int v=e[i].v;
if(v==pre && !k){
k=1;
continue ;
}
if(!dfn[v]){
Tarjan(v,u);
low[u]=min(low[u],low[v]);
}else{
low[u]=min(low[u],dfn[v]);
}
if(low[v]>dfn[u]){
cut[v]=1;
}else Union(u,v);
}
}
int mx=0,node=1;
void dfs(int u,int d,int pa){
if(d>mx){
mx=d;
node=u;
}
for(int i=head[u];i!=-1;i=e[i].next){
int v=e[i].v;
if(v==pa) continue ;
dfs(v,d+1,u);
}
}
int main(){
while(scanf("%d%d",&n,&m)!=EOF){
if(!n&&!m) break ;
init();
for(int i=1;i<=m;i++){
int u,v;
scanf("%d%d",&u,&v);
adde(u,v);adde(v,u);
if(u>v)swap(u,v);
g[i].u=u;g[i].v=v;
}
sort(g+1,g+m+1);
Tarjan(1,0);
memset(head,-1,sizeof(head));tot=0;
for(int i=1;i<=m;i++){
int u=g[i].u,v=g[i].v;
if(g[i].u==g[i-1].u&&g[i].v==g[i-1].v) continue ;
int fx=find(u),fy=find(v);
if(!num[fx]) num[fx]=++cnt;
if(!num[fy]) num[fy]=++cnt;
if(num[fx]==num[fy]) continue ;
adde(num[fx],num[fy]);adde(num[fy],num[fx]);
}
mx=0;
dfs(1,0,-1);
mx=0;
dfs(node,0,-1);
cout<<cnt-1-mx<<endl;
}
return 0;
}
原文地址：https://www.cnblogs.com/heyuhhh/p/10393312.html