Warm up
Time Limit:5000MS Memory Limit:65535KB 64bit IO Format:%I64d & %I64u
Submit Status Practice HDU 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
题目大意：有n颗行星，有m条双向通道连接着m对行星。问你新建一条双向通道后，无向图中最少会剩下多少条桥。有重边。
解题思路：无向图求边双连通分量，缩点，重新构图，形成树。求树的直径，然后用原图总的桥减去树的直径即为结果。求树的直径，我们用两次搜索，第一次从任意点出发，搜到的最远结点即为直径的一端，然后从这一端再次进行搜索，搜到直径的另一端。
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<vector>
#include<stack>
using namespace std;
const int maxn = 200100;
struct Edge{
int from,to,dist,next;
Edge(){}
Edge(int _to,int _next):to(_to),next(_next){}
}edges[maxn*10];
int head[maxn], tot;
int dfs_clock, dfn[maxn], brinum;
int Stack[maxn], instack[maxn], top, ebccno[maxn], ebcc_cnt;
int deg[maxn];
vector<int>G[maxn];
void init(){
tot = 0;
brinum = dfs_clock = 0;
top = 0;
ebcc_cnt = 0;
memset(deg,0,sizeof(deg));
memset(head,-1,sizeof(head));
}
void AddEdge(int _u,int _v){
edges[tot] = Edge(_v,head[_u]);
head[_u] = tot++;
}
int dfs(int u,int fa){
int lowu = dfn[u] = ++dfs_clock;
Stack[++top] = u;
// instack[u] = 1;
for(int i = head[u]; i != -1; i = edges[i].next){
int v = edges[i].to;
if(!dfn[v]){
int lowv = dfs(v,i);
lowu = min(lowu,lowv);
if(lowv > dfn[u]){
brinum++;
}
}else if(dfn[v] < dfn[u] && (fa^1) != i){//这里用边的编号来标记是否是同一条边的回边
lowu = min(lowu,dfn[v]);
}
}
if(dfn[u] == lowu){ //找到一个边双连通分量
ebcc_cnt++;
for(;;){
int v = Stack[top--];
// instack[v] = 0;
ebccno[v] = ebcc_cnt; //给每个点划分一个分量标号
if(u == v){
break;
}
}
}
// low[u] = lowu;
return lowu;
}
void find_ebcc(int n){
memset(dfn,0,sizeof(dfn));
memset(instack,0,sizeof(instack));
for(int i = 1; i <= n; i++){
if(!dfn[i]){
dfs(i,-1);
}
}
}
int pos, Maxd;
void dfs1(int u,int dep,int fa){ //求树的直径
if(dep > Maxd){
Maxd = dep;
pos = u;
}
for(int i = 0; i < G[u].size(); i++){
int v = G[u][i];
if(fa == v){ continue; }
dfs1(v,dep+1,u);
}
}
int main(){
int n,m;
while(scanf("%d%d",&n,&m)!=EOF&&(n+m)){
init();
for(int i = 0; i <= n; i++){
G[i].clear();
}
int a,b;
for(int i = 0; i < m; i++){
scanf("%d%d",&a,&b);
AddEdge(a,b);
AddEdge(b,a);
}
find_ebcc(n);
for(int i = 1; i <= n; i++){
for(int j = head[i]; j != -1; j = edges[j].next){
int v = edges[j].to;
if(ebccno[i] != ebccno[v]){ //重新构图，形成树
G[ebccno[i]].push_back(ebccno[v]);
}
}
}
pos = 1, Maxd = 0;
dfs1(1,0,-1);
int st = pos; Maxd = 0;
dfs1(pos,0,-1);
printf("%d\n",brinum - Maxd);
}
return 0;
}