【连通图|边双连通+缩点】POJ-3352 Road Construction

Road Construction

Time Limit: 2000MS Memory Limit: 65536K

Description

It’s almost summer time, and that means that it’s almost summer construction time! This year, the good people who are in charge of the roads on the tropical island paradise of Remote Island would like to repair and upgrade the various roads that lead between the various tourist attractions on the island.

The roads themselves are also rather interesting. Due to the strange customs of the island, the roads are arranged so that they never meet at intersections, but rather pass over or under each other using bridges and tunnels. In this way, each road runs between two specific tourist attractions, so that the tourists do not become irreparably lost.

Unfortunately, given the nature of the repairs and upgrades needed on each road, when the construction company works on a particular road, it is unusable in either direction. This could cause a problem if it becomes impossible to travel between two tourist attractions, even if the construction company works on only one road at any particular time.

So, the Road Department of Remote Island has decided to call upon your consulting services to help remedy this problem. It has been decided that new roads will have to be built between the various attractions in such a way that in the final configuration, if any one road is undergoing construction, it would still be possible to travel between any two tourist attractions using the remaining roads. Your task is to find the minimum number of new roads necessary.

Input

The first line of input will consist of positive integers n and r, separated by a space, where 3 ≤ n ≤ 1000 is the number of tourist attractions on the island, and 2 ≤ r ≤ 1000 is the number of roads. The tourist attractions are conveniently labelled from 1 to n. Each of the following r lines will consist of two integers, v and w, separated by a space, indicating that a road exists between the attractions labelled v and w. Note that you may travel in either direction down each road, and any pair of tourist attractions will have at most one road directly between them. Also, you are assured that in the current configuration, it is possible to travel between any two tourist attractions.

Output

One line, consisting of an integer, which gives the minimum number of roads that we need to add.

Sample Input

Sample Input 1

10 12

1 2

1 3

1 4

2 5

2 6

5 6

3 7

3 8

7 8

4 9

4 10

9 10

Sample Input 2

3 3

1 2

2 3

1 3

Sample Output

Output for Sample Input 1

2

Output for Sample Input 2

0

Source

CCC 2007

思路： 一开始我还以为是要求边连通度，但是边连通度的意义并不是这样。边连通度指的是至少删除多少条边可以使得原来的连通图变得不连通。后来发现原来和这道题是一模一样的：参考POJ-3177

统计出树中度为1的节点的个数，即为叶节点的个数，记为leaf。则至少在树上添加(leaf+1)/2条边，就能使树达到边双连通，所以至少添加的边数就是(leaf+1)/2。

代码如下：

/*
 * ID: j.sure.1
 * PROG:
 * LANG: C++
 */
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <ctime>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <map>
#include <set>
#include <string>
#include <climits>
#include <iostream>
#define PB push_back
#define LL long long
using namespace std;
const int INF = 0x3f3f3f3f;
const double eps = 1e-8;
/****************************************/
const int N = 1e3 + 5, M = 2e3 + 5;
struct Edge {
    int v, next, idx;
    Edge(){}
    Edge(int _v, int _next, int _idx):
        v(_v), next(_next), idx(_idx){}
}e[M];
int n, m, deep, tot, bcc_cnt;
int dfn[N], head[N], bcc_id[N], deg[N], line[N][2];
bool isbri[N];

void init()
{
    bcc_cnt = tot = deep = 0;
    memset(head, -1, sizeof(head));
    memset(isbri, 0, sizeof(isbri));
    memset(deg, 0, sizeof(deg));
    memset(dfn, 0, sizeof(dfn));
    memset(bcc_id, 0, sizeof(bcc_id));
}

void add(int u, int v, int idx)
{
    e[tot] = Edge(v, head[u],idx);
    head[u] = tot++;
}

int dfs(int u, int fa)
{
    int lowu = dfn[u] = ++deep;
    for(int i = head[u]; ~i; i = e[i].next) {
        int v = e[i].v;
        if(!dfn[v]) {
            int lowv = dfs(v, u);
            lowu = min(lowu, lowv);
            if(lowv > dfn[u]) isbri[e[i].idx] = true;
        }
        else if(dfn[v] < dfn[u] && v != fa) lowu = min(lowu, dfn[v]);
    }
    return lowu;
}

void flood(int u)
{
    bcc_id[u] = bcc_cnt;
    for(int i = head[u]; ~i; i = e[i].next) {
        if(isbri[e[i].idx]) continue;
        int v = e[i].v;
        if(!bcc_id[v]) flood(v);
    }
}

int main()
{
#ifdef J_Sure
    freopen("000.in", "r", stdin);
    //freopen("999.out", "w", stdout);
#endif
    scanf("%d%d", &n, &m);
    init();
    int u, v, ID = 0;
    for(int i = 0; i < m; i++) {
        scanf("%d%d", &u, &v);
        u--; v--;
        line[i][0] = u; line[i][1] = v;
        add(u, v, ID);
        add(v, u, ID++);
    }
    dfs(0, -1);
    for(int i = 0; i < n; i++) {
        if(!bcc_id[i]) {
            bcc_cnt++;
            flood(i);
        }
    }
    for(int i = 0; i < m; i++) {
        int u = bcc_id[line[i][0]], v = bcc_id[line[i][1]];
        if(u != v) {
            deg[u]++; deg[v]++;
        }
    }
    int ans = 0;
    for(int i = 1; i <= bcc_cnt; i++) {
        if(deg[i] == 1) ans++;
    }
    printf("%d\n", (ans+1) / 2);
    return 0;
}