?pid=4612
Problem 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 41 21 31 42 30 0
Sample Output
0
Source
/**hdu4612 连通性，求树的直径，，加一边求最少桥题目大意：给定一个无向连通图加上一条边后所得到的图所含的桥的数目最少解题思路：tarjan缩点边树，求出树的直径m，则原有桥数-m即为所求。本题要注意考虑重边和注意求树的直径的方法（见代码注释）*直径:；任意两个节点之间的最长距离*/#pragma comment(linker, "/STACK:1024000000,1024000000")///申请空间#include <stdio.h>#include <string.h>#include <algorithm>#include <iostream>#include <vector>using namespace std;const int maxn=200010;const int maxm=2000014;struct note{int v,next;bool cut,chong_bian;} edge[maxm];int head[maxn],ip;int n,m;void init(){memset(head,-1,sizeof(head));ip=0;}void addedge(int u,int v,bool chong_bian){edge[ip].v=v,edge[ip].cut=false,edge[ip].chong_bian=chong_bian,edge[ip].next=head[u],head[u]=ip++;}int dfn[maxn],low[maxn],dex,inst[maxn],st[maxn],top,cnt,belong[maxn];void tarjan(int u,int pre,bool ff)///ff代表重边{dfn[u]=low[u]=++dex;st[top++]=u;inst[u]=1;for(int i=head[u]; i!=-1; i=edge[i].next){int v=edge[i].v;if(v==pre&&(!ff))continue;///与没有重边的区别if(!dfn[v]){tarjan(v,u,edge[i].chong_bian);if(low[u]>low[v])low[u]=low[v];if(low[v]>dfn[u]){///bridge++;edge[i].cut=true;edge[i^1].cut=true;}}else if(inst[v]&&dfn[v]<low[u]){low[u]=dfn[v];}}if(dfn[u]==low[u]){cnt++;int v;do{v=st[–top];inst[v]=0;belong[v]=cnt;}while(v!=u);}}vector <int> vec[maxn];int dep[maxn];void dfs(int u)///dfs求每个节点的深度{for(int i=0; i<vec[u].size(); i++){int v=vec[u][i];if(dep[v]!=-1)continue;dep[v]=dep[u]+1;dfs(v);}}void solve(){memset(dfn,0,sizeof(dfn));memset(inst,0,sizeof(inst));cnt=dex=top=0;tarjan(1,0,false);for(int i=1;i<=cnt;i++)vec[i].clear();for(int u=1; u<=n; u++){for(int i=head[u]; i!=-1; i=edge[i].next){if(edge[i].cut){int v=edge[i].v;vec[belong[u]].push_back(belong[v]);}}}///=======求直径=========memset(dep,-1,sizeof(dep));dep[1]=0;dfs(1);int k=1;for(int i=1; i<=cnt; i++){if(dep[i]>dep[k])k=i;}memset(dep,-1,sizeof(dep));dep[k]=0;dfs(k);int ans=0;for(int i=1; i<=cnt; i++){ans=max(ans,dep[i]);}///=======================printf("%d\n",cnt-1-ans);}struct note_{int x,y;} node[maxm];bool cmp(note_ a,note_ b){if(a.x==b.x)return a.y<b.y;return a.x<b.x;}int main(){while(~scanf("%d%d",&n,&m)){if(n==0&&m==0)break;for(int i=0; i<m; i++){int u,v;scanf("%d%d",&u,&v);if(u==v)continue;if(u>v)swap(u,v);node[i].x=u;node[i].y=v;}sort(node,node+m,cmp);init();for(int i=0; i<m; i++){if(i==0||(node[i].x!=node[i-1].x||node[i].y!=node[i-1].y)){if(i<m-1&&node[i].x==node[i+1].x&&node[i].y==node[i+1].y){addedge(node[i].x,node[i].y,true);addedge(node[i].y,node[i].x,true);}else{addedge(node[i].x,node[i].y,false);addedge(node[i].y,node[i].x,false);}}}solve();}return 0;}
因为在路上你就已经收获了自由自在的好心情。
