Road Construction

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

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

3 3
1 2
2 3
1 3

Sample Output

2
0
题意：给出一个连通图，求至少添加几条边，使得图去掉任意一条边，还是连通的。
强连通分量缩点生成一颗树，求入度为1的连通分量个数n，结果为(n+1)/2；
无向图不用像有向图那样考虑是否该点在栈中，因为u跟v只要有边，说明u一定可以到到达v，v可以到达u；
而有向图，比如1->2, 3->2,假如从1搜索到2，1跟2属于两个独立的强连通分量，而从3搜到2，发现2被搜过，但2不在栈里，所以
不能使low[3] = low[2];所以无向图的tarjan比有向图的简单了很多。

代码：


#include <stdio.h>
#include <stdlib.h>
#define Min(a, b) a < b ? a : b
#define Max(a, b) a > b ? a : b
#define maxn 1001
struct edge
{
    int v, next;
}fn[maxn * maxn];
int visit[maxn], g[maxn], du[maxn],  low[maxn];
void set(int n)
{
    for (int i = 1; i <= n; i++)
    {
        visit[i] = 0, g[i] = -1, du[i] = 0;
    }
}
int tarjan(int f, int u, int times)
{
    visit[u] = 1;
    low[u] = times;
    int i, v;
    for (i = g[u]; i != -1; i = fn[i].next)
    {
        v = fn[i].v;
        if (v != f)
        {
            if (!visit[v])
            {
                times = tarjan(u, v, times + 1);
                low[u] = Min(low[u], low[v]);
            }
            low[u] = Min(low[u], low[v]);
        }
    }
    return times;
}
int main()
{
    char a[10], b[10];
    int ca, n, m, th, u, v, i, times, top, k, j;
    while (scanf("%d%d", &n, &m) != EOF)
    {
        set(n);
        th = 0;
        while (m--)
        {
            scanf("%d%d", &u, &v);
            fn[th].v = v, fn[th].next = g[u], g[u] = th++;
            fn[th].v = u, fn[th].next = g[v], g[v] = th++;
        }
        for (i = 1, times = 0; i <= n; i++)
        {
            if (!visit[i])
            {
                times = tarjan(-1, i, times + 1); 
            }
        }
        for (i = 1; i <= n; i++)
        {
            for (j = g[i]; j != -1; j = fn[j].next)
            {
                if (low[i] != low[fn[j].v])
                {
                    du[low[i]]++;
                }
            }
        }
        for (i = 1, k = 0; i <= n; i++)
        {
            if (du[i] == 1)
            {
                k++;
            }
        }
        printf("%d\n", (k + 1) / 2);
    }
    system("pause");
    return 0;
}

代码2：


#include <stdio.h>
#include <stdlib.h>
#define Min(a, b) a < b ? a : b
#define Max(a, b) a > b ? a : b
#define maxn 1001
struct edge
{
    int v, next;
}fn[maxn * maxn];
int visit[maxn], scc[maxn], g[maxn], du[maxn], stack[maxn], dfn[maxn], low[maxn], top, num, hash[maxn];
void set(int n)
{
    for (int i = 1; i <= n; i++)
    {
        visit[i] = 0, scc[i] = -1, g[i] = -1, du[i] = 0, hash[i] = 0;
    }
}
int tarjan(int f, int u, int times)
{
    visit[u] = 1;
    stack[top++] = u;
    dfn[u] = low[u] = times;
    int i, v;
    for (i = g[u]; i != -1; i = fn[i].next)
    {
        v = fn[i].v;
        if (v != f)
        {
            if (!visit[v])
            {
                times = tarjan(u, v, times + 1);
                low[u] = Min(low[u], low[v]);
            }
            else if (scc[v] == -1)
            {
                low[u] = Min(low[u], low[v]);
            }
        }
    }
    if (low[u] == dfn[u])
    {
        num++;
        do
        {
            scc[stack[--top]] = num;
            //printf("%d ", stack[top]);
        }while (stack[top] != u);//printf("-=-=\n");
    }
    return times;
}
int main()
{
    char a[10], b[10];
    int ca, n, m, th, u, v, i, times, top, k, j;
    scanf("%d%d", &n, &m);
        set(n);
        th = 0;
        while (m--)
        {
            scanf("%d%d", &u, &v);
            fn[th].v = v, fn[th].next = g[u], g[u] = th++;
            fn[th].v = u, fn[th].next = g[v], g[v] = th++;
        }
        for (i = 1, times = 0, top = 0, num = 0; i <= n; i++)
        {
            if (!visit[i])
            {
                times = tarjan(-1, i, times); 
            }
        }
        for (i = 1; i <= n; i++)
        {
            hash[i] = 1;
            for (j = g[i]; j != -1; j = fn[j].next)
            {
                if (scc[i] != scc[fn[j].v])
                {
                    du[scc[i]]++;
                }
            }
        }
        for (i = 1, k = 0; i <= num; i++)
        {
            //printf("[%d] = %d ", i, du[i]);
            if (du[i] == 1)
            {
                k++;
            }
        }
        printf("%d\n", (k + 1) / 2);
    return 0;
}