hdu 4612 Warm up 桥缩点
题目:给出一个图,添加一条边之后,问能够在新图中得到的最少的桥的数量。
分析:我们可以双联通分量进行缩点,原图变成了一棵树。问题变成了:求树中添加一条边之后,使得不在圈的边最少。显然求一边直径,用总边数减掉最长路上的边数就是答案。注意数据存在重边的情况。
#include <set>
#include <map>
#include <list>
#include <cmath>
#include <queue>
#include <stack>
#include <string>
#include <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define debug puts("here")
#define rep(i,n) for(int i=0;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define REP(i,a,b) for(int i=a;i<=b;i++)
#define foreach(i,vec) for(unsigned i=0;i<vec.size();i++)
#define pb push_back
#define RD(n) scanf("%d",&n)
#define RD2(x,y) scanf("%d%d",&x,&y)
#define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define RD4(x,y,z,w) scanf("%d%d%d%d",&x,&y,&z,&w)
#define All(vec) vec.begin(),vec.end()
#define MP make_pair
#define PII pair<int,int>
#define PQ priority_queue
#define cmax(x,y) x = max(x,y)
#define cmin(x,y) x = min(x,y)
#define Clear(x) memset(x,0,sizeof(x))
/*
#pragma comment(linker, "/STACK:1024000000,1024000000")
int size = 256 << 20; // 256MB
char *p = (char*)malloc(size) + size;
__asm__("movl %0, %%esp\n" :: "r"(p) );
*/
char IN;
bool NEG;
inline void Int(int &x){
NEG = 0;
while(!isdigit(IN=getchar()))
if(IN=='-')NEG = 1;
x = IN-'0';
while(isdigit(IN=getchar()))
x = x*10+IN-'0';
if(NEG)x = -x;
}
inline void LL(ll &x){
NEG = 0;
while(!isdigit(IN=getchar()))
if(IN=='-')NEG = 1;
x = IN-'0';
while(isdigit(IN=getchar()))
x = x*10+IN-'0';
if(NEG)x = -x;
}
/******** program ********************/
const int MAXN = 200005;
const int MAXM = 1000005;
vector<int> adj[MAXN];
int po[MAXN],tol;
int low[MAXN],dfn[MAXN],tim;
bool use[MAXN];
int fa[MAXN];
int id[MAXN],ha[MAXN];
struct Edge{
int y,id,next;
}edge[MAXM<<1];
inline void add(int x,int y,int i){
edge[++tol].y = y;
edge[tol].id = i;
edge[tol].next = po[x];
po[x] = tol;
}
int findSet(int x){
if(x!=fa[x])
fa[x] = findSet(fa[x]);
return fa[x];
}
void dfs(int x,int fid){
low[x] = dfn[x] = ++ tim;
for(int i=po[x];i;i=edge[i].next){
int y = edge[i].y;
int id = edge[i].id;
if(id==fid)continue;
if(!dfn[y]){
dfs(y,id);
cmin( low[x],low[y] );
if(low[y]<=dfn[x]){
int px = findSet(x);
int py = findSet(y);
if(px!=py)fa[px] = py;
}
}else
cmin( low[x],dfn[y] );
}
}
int MAX,root;
void dfsR(int x,int sz){
use[x] = true;
if(sz>MAX){
MAX = sz;
root = x;
}
foreach(i,adj[x])
if(!use[adj[x][i]])
dfsR(adj[x][i],sz+1);
}
int main(){
#ifndef ONLINE_JUDGE
freopen("1002.in","r",stdin);
//freopen("sum.out","w",stdout);
#endif
int size = 256 << 20; // 256MB
char *p = (char*)malloc(size) + size;
__asm__("movl %0, %%esp\n" :: "r"(p) );
int x,y,n,m;
while(1){
Int(n);Int(m);
if(!n&&!m)break;
Clear(po);
tol = 0;
rep1(i,m){
Int(x);Int(y);
add(x,y,i);
add(y,x,i);
}
rep1(i,n){
adj[i].clear();
fa[i] = i;
}
Clear(dfn);
tim = 0;
rep1(i,n)
if(!dfn[i])
dfs(i,0);
Clear(id);
int tot = 0;
rep1(i,n){
int px = findSet(i);
if(!id[px])id[px] = ++tot;
ha[i] = id[px];
}
rep1(x,n){
int px = ha[x];
for(int i=po[x];i;i=edge[i].next){
int py = ha[ edge[i].y ];
if(px==py)continue;
adj[px].pb(py);
adj[py].pb(px);
}
}
Clear(use);
MAX = 0;
dfsR(1,1);
Clear(use);
MAX = 0;
dfsR(root,1);
printf("%d\n",tot-MAX);
}
return 0;
}

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