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 71 22 33 42 54 55 65 7
```

Sample Output

```				2题意：给出一个连通图，求至少添加多少条边，使得对于任意两点，不只一条路。即两点间的路去掉一条边还是连通的。
#include <stdio.h>#include <stdlib.h>#define Min(a, b) a < b ? a : b#define maxn 5001struct T{    int v, next;}fn[maxn * 4];int g[maxn], indegree[maxn], visit[maxn], low[maxn];void set(int n){    for (int i = 1; i <= n; i++)    {        g[i] = -1;        indegree[i] = 0;        visit[i] = 0;    }    }int tarjan(int u, int f, int time){    int i, v;    visit[u] = 1;    low[u] = time;    for (i = g[u]; i != -1; i = fn[i].next)    {        v = fn[i].v;        if (v != f)        {            if(!visit[v])            {                time = tarjan(v, u, time + 1);            }            low[u] = Min(low[u], low[v]);        }    }    return time;}int isok(int u, int v){    int i;    for (i = g[u]; i != -1; i = fn[i].next)    {        if (fn[i].v == v)        {            //printf("ooo\n");            return 0;        }    }    return 1;}int main(){    int n, m, u, v, th, sum, i, j;    while (scanf("%d%d", &n, &m) != EOF)    {        th = 0;        set(n);        while (m--)        {            scanf("%d%d", &u, &v);            if (isok(u,v))//处理重边，重边在求双连通分量时没影响，但在统计度时，由于重边也要重建，所以度会变多             {                fn[th].v = v, fn[th].next = g[u], g[u] = th++;                fn[th].v = u, fn[th].next = g[v], g[v] = th++;            }        }        tarjan(1, -1, 1);        for (i = 1; i <= n; i++)        {            for (j = g[i]; j != -1; j = fn[j].next)            {                if (low[i] != low[fn[j].v])                {                    indegree[low[i]]++;                }            }        }        for (i = 1, sum = 0; i <= n; i++)        {            if (indegree[i] == 1)            {                sum++;            }        }        printf("%d\n", (sum + 1) / 2);    }    //system("pause");    return 0;}/*首先这道题有重边，如2 21 21 2应该输出 12 21 22 1*/

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