poj3177 && poj3352 边双连通分量缩点

Redundant Paths

Description

In order to get from one of the F (1 <= F <= 5,000) grazing fields (which are numbered 1..F) to another field, Bessie and the rest of the herd are forced to cross near the Tree of Rotten Apples. The cows are now tired of often being forced to take a particular path and want to build some new paths so that they will always have a choice of at least two separate routes between any pair of fields. They currently have at least one route between each pair of fields and want to have at least two. Of course, they can only travel on Official Paths when they move from one field to another.

Given a description of the current set of R (F-1 <= R <= 10,000) paths that each connect exactly two different fields, determine the minimum number of new paths (each of which connects exactly two fields) that must be built so that there are at least two separate routes between any pair of fields. Routes are considered separate if they use none of the same paths, even if they visit the same intermediate field along the way.

There might already be more than one paths between the same pair of fields, and you may also build a new path that connects the same fields as some other path.

Input

Line 1: Two space-separated integers: F and R

Lines 2..R+1: Each line contains two space-separated integers which are the fields at the endpoints of some path.

Output

Line 1: A single integer that is the number of new paths that must be built.

Sample Input

7 7
1 2
2 3
3 4
2 5
4 5
5 6
5 7

Sample Output

2

题意:

给你一张无向图，判断至少要加多少边，才能使任意2个点间至少有2条相互独立（无公共边）的道路；

思路:

在同一个双连通分量中的点可以等价于一个点，原图就成了一棵树，那么问题就转化为树中加多少条边可以成为双连通图。答案 = （树中度为1的边个数 + 1） / 2；

/*
 * Author:  sweat123
 * Created Time:  2016/6/21 20:07:00
 * File Name: main.cpp
 */
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<string>
#include<vector>
#include<cstdio>
#include<time.h>
#include<cstring>
#include<iostream>
#include<algorithm>
#define INF 1<<30
#define MOD 1000000007
#define ll long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define pi acos(-1.0)
using namespace std;
const int MAXN = 5010;
struct node{
    int to;
    int next;
}edge[MAXN<<2];
int pre[MAXN],vis[MAXN],pa[MAXN],dfn[MAXN],low[MAXN],n,m,ind;
int px[MAXN],py[MAXN];
int pcnt;
void add(int x,int y){
    edge[ind].to = y;
    edge[ind].next = pre[x];
    pre[x] = ind ++;
}
int find(int x){
    if(pa[x] != x)pa[x] = find(pa[x]);
    return pa[x];
}
void dfs(int rt,int k,int fa){
    dfn[rt] = low[rt] = k;
    for(int i = pre[rt]; i != -1; i = edge[i].next){
        int t = edge[i].to;
        if(!dfn[t] && t != fa){
            dfs(t,k+1,rt);
            low[rt] = min(low[rt],low[t]);
            if(low[t] > dfn[rt]){
                px[pcnt] = rt,py[pcnt++] = t;
            } else {
                int fx = find(t);
                int fy = find(rt);
                if(pa[fx] != pa[fy]){
                    pa[fx] = fy;
                }
            }
        } else if(t != fa){//bridge is differenet from point
            low[rt] = min(low[rt],dfn[t]);
        }
    }
}
int d[MAXN],f[MAXN];
int main(){
    while(~scanf("%d%d",&n,&m)){
        ind = 0;
        pcnt = 0;
        memset(pre,-1,sizeof(pre));
        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++){
            pa[i] = i;
        }
        memset(dfn,0,sizeof(dfn));
        memset(low,0,sizeof(low));
        dfs(1,1,-1);
        //for(int i = 1; i <= n; i++){
            //cout<<dfn[i]<<‘ ‘<<low[i]<<endl;
        //}
        //cout<<endl;
        memset(d,0,sizeof(d));
        memset(f,-1,sizeof(f));
        int pnum = 0;
        for(int i = 1; i <= n; i++){
            int fx = find(i);
            if(f[fx] == -1)f[fx] = ++pnum;
            f[i] = f[fx];
        }
        for(int i = 0; i < pcnt; i++){
            d[f[px[i]]] ++,d[f[py[i]]] ++;
        }
        int ans = 0;
        for(int i = 1; i <= pnum; i++){
            if(d[i] == 1)ans ++;
        }
        printf("%d\n",(ans + 1) / 2);
    }
    return 0;
}