CodeForces 25D Roads not only in Berland【并查集】

D. Roads not only in Berland

time limit per test

 2 seconds

memory limit per test

 256 megabytes

input

 standard input

output

 standard output

Berland Government decided to improve relations with neighboring countries. First of all, it was decided to build new roads so that from each city of Berland and neighboring countries it became possible to reach all the others. There are n cities in Berland and neighboring countries in total and exactly n - 1 two-way roads. Because of the recent financial crisis, the Berland Government is strongly pressed for money, so to build a new road it has to close some of the existing ones. Every day it is possible to close one existing road and immediately build a new one. Your task is to determine how many days would be needed to rebuild roads so that from each city it became possible to reach all the others, and to draw a plan of closure of old roads and building of new ones.

Input

The first line contains integer n (2 ≤ n ≤ 1000) — amount of cities in Berland and neighboring countries. Next n - 1 lines contain the description of roads. Each road is described by two space-separated integers ai, bi (1 ≤ ai, bi ≤ n, ai ≠ bi) — pair of cities, which the road connects. It can't be more than one road between a pair of cities. No road connects the city with itself.

Output

Output the answer, number t — what is the least amount of days needed to rebuild roads so that from each city it became possible to reach all the others. Then output t lines — the plan of closure of old roads and building of new ones. Each line should describe one day in the format i j u v — it means that road between cities i and j became closed and a new road between cities u and v is built. Cities are numbered from 1. If the answer is not unique, output any.

Sample test(s)

input

2
1 2

output

0

input

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

output

1
3 1 3 7

 题目大意：每一天都能拆一条路，并且重新建造一条路，问需要多少天，才能使得所有道路连通在一起、

题目解析：自环的部分就是要拆除的路，至于需要多少天，那么就要看有多少条分开的路，比如题干中样例2、

拆3 1那条路是一定的，然后这个时候分出来两个路：1 2 3是一条3 4 5 6 是一条（也可以说是在同一个集和里边）

然后我们的任务就是连接两条路，合并成一条就行了~

#include
#include
using namespace std;
int u[313131];
int v[313131];
int pos[121212];
int f[3131313];
int s[1212121];
int find(int x)
{
    return f[x] == x ? x : (f[x] = find(f[x]));
}
void merge(int a,int b)
{
    int A,B;
    A=find(a);
    B=find(b);
    if(A!=B)
    f[B]=A;
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        int cont=0;
        for(int i=1;i<=n;i++)
        {
            f[i]=i;
        }
        for(int i=0;i