hdu 4612 强连通

题意:有一些联通的地方，如果2点间只有一条路径，这样的边叫做桥，现在让你添加一个桥，使最后的桥最少，问最少的桥使多少？

　　先求一次强连通分量，然后图就分成了几个块，将这几个块看做点，求出总共有多少条重建图中的边，然后再求出最长的边，这样答案就是总共

边数 - 最长的边的长度。原来的标记手法过了几题，但是做这题有点问题，用了另一个手法。

#pragma comment(linker, "/STACK:102400000,102400000")
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define INF 1000000001
#define ll __int64
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
const int MAXN = 200010;
struct node
{
int to;
int next;
int vis;
}edge[1000010 * 2],a[1000010 * 2];
struct fnode
{
int x;
int val;
};
int n,m,ind,pre[MAXN],ins[MAXN],low[MAXN],dfn[MAXN],vis[MAXN],num[MAXN],fpre[MAXN];
int fp,len,p;
stack<int>s;
void Init()
{
fp = 1;
ind = 0;
while(!s.empty())s.pop();
memset(num,0,sizeof(num));
memset(ins,0,sizeof(ins));
memset(pre,-1,sizeof(pre));
memset(vis,0,sizeof(vis));
memset(dfn,0,sizeof(dfn));
memset(low,0,sizeof(low));
}
void add(int x,int y)
{
edge[ind].to = y;
edge[ind].vis = 0;
edge[ind].next = pre[x];
pre[x] = ind ++;
}
void dfs(int rt,int k,int fa)
{
s.push(rt);
vis[rt] = ins[rt] = 1;
low[rt] = dfn[rt] = k;
for(int i = pre[rt]; i != -1; i = edge[i].next){
int t = edge[i].to;
if(edge[i].vis)continue;
edge[i].vis = edge[i^1].vis = 1;
//原来判断 fa != t 但是有重边的情况这种标记不行。
if(!dfn[t]){
dfs(t,k+1,rt);
low[rt] = min(low[rt],low[t]);
}
else{
low[rt] = min(dfn[t],low[rt]);
}
}
if(low[rt] == dfn[rt]){
while(!s.empty()){
int temp = s.top();
s.pop();
num[temp] = fp;
if(temp == rt)break;
}
fp ++;
}
}
void add_b(int x,int y)
{
a[ind].to = y;
a[ind].next = fpre[x];
fpre[x] = ind ++;
}
void build(int rt)
{
vis[rt] = 1;
for(int i = pre[rt]; i != -1; i=edge[i].next){
int t = edge[i].to;
if(!vis[t]){
if(num[t] != num[rt]){
add_b(num[t],num[rt]);
add_b(num[rt],num[t]);
}
build(t);
}
}
}
void bfs(int rt)
{
queue<fnode>q;
fnode temp;
temp.val = 0;
temp.x = rt;
vis[rt] = 1;
q.push(temp);
while(!q.empty()){
temp = q.front();
q.pop();
for(int i = fpre[temp.x]; i != -1; i = a[i].next){
int t = a[i].to;
if(!vis[t]){
fnode ft;
ft.val = temp.val + 1;
ft.x = t;
vis[t] = 1;
q.push(ft);
if(ft.val > len){
len = ft.val;
p = ft.x;
}
}
}
}
}
int main()
{
while(~scanf("%d%d",&n,&m)){
if(!n && !m)break;
Init();
for(int i = 1; i <= m; i++){
int x,y;
scanf("%d%d",&x,&y);
add(x,y);
add(y,x);
}
for(int i = 1; i <= n; i++){
if(!dfn[i]){
dfs(i,1,-1);
}
}
memset(vis,0,sizeof(vis));
ind = 1;
memset(fpre,-1,sizeof(fpre));
for(int i = 1; i <= n; i++){
if(!vis[i]){
build(i);
}
}
len = 0;
p = 1;
memset(vis,0,sizeof(vis));
bfs(1);
len = 0;
memset(vis,0,sizeof(vis));
bfs(p);
fp -= 1;
int ans = fp - 1 - len;
//cout<<"fp = "<<fp<<" len = "<<len<<endl;
cout<<(ans > 0 ? ans : 0)<<endl;
}
return 0;
}