ZOJ 3321 Circle(并查集啊)

标签：zoj   并查集   

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=3731

Your task is so easy. I will give you an undirected graph, and you just need to tell me whether the graph is just a circle. A cycle is three or more nodes V₁, V₂, V₃, ...V_k, such that there are edges between V₁ and V₂, V₂ and V₃, ... V_k and V₁, with no other extra edges. The graph will not contain self-loop. Furthermore, there is at most one edge between two nodes.

Input

There are multiple cases (no more than 10).

The first line contains two integers n and m, which indicate the number of nodes and the number of edges (1 < n < 10, 1 <= m < 20).

Following are m lines, each contains two integers x and y (1 <= x, y <= n, x != y), which means there is an edge between node x and node y.

There is a blank line between cases.

Output

If the graph is just a circle, output "YES", otherwise output "NO".

Sample Input

3 3
1 2
2 3
1 3

4 4
1 2
2 3
3 1
1 4

Sample Output

YES
NO

题意：

求所给的所有边能恰好形成一个圆！

代码一：

#include <cstdio>
#include <cstring>
int f[147];
int ma[147];
int find(int x)
{
    return x==f[x] ? x:f[x]=find(f[x]);
}

void init(int n)
{
    for(int i = 0; i <= n; i++)
    {
        f[i] = i;
    }
}
void Union(int x, int y)
{
    int f1 = find(x);
    int f2 = find(y);
    if(f1 != f2)
    {
        f[f2] = f1;
    }
}

int main()
{
    int n, m;
    while(~scanf("%d%d",&n,&m))
    {
        memset(ma, 0,sizeof(ma));
        int a, b;
        init(n);
        for(int i = 0; i < m; i++)
        {
            scanf("%d%d",&a,&b);
            Union(a,b);
            ma[a]++, ma[b]++;
        }
        int flag = 0;
        for(int i = 1; i <= n; i++)
        {
            if(ma[i] != 2)//每个点只能出现两次
            {
                flag = 1;
                break;
            }
            if(f[i] != f[n])
            {
                flag = 1;
                break;
            }
        }
        if(flag)
        {
            printf("NO\n");
        }
        else
        {
            printf("YES\n");
        }
    }
    return 0;
}

#include <cstdio>
#include <cstring>
int main()
{
    int n, m;
    int map[25][25];
    while(scanf("%d%d", &n, &m)!=EOF)
    {
        int a, b;
        memset(map, 0, sizeof(map));
        for(int i = 0; i < m; i++)
        {
            scanf("%d%d", &a, &b);
            map[a][b] = map[b][a] = 1;
        }
        if(n != m)
        {
            printf("NO\n");
            continue;
        }
        int ans = 0;
        int flag = 0;
        int tt = 1;
        for(int i = 1; i <= m; i++)
        {
            flag = 0;
            for(int j = 1; j <= n; j++)//从点1开始循环，如果最后可以回到点1，就说明是一个圆
                if(map[tt][j])
                {
                    map[tt][j] = map[j][tt] = 0;//标记已找到的边
                    tt = j;
                    flag = 1;
                    break;
                }
            if(flag == 0)//没找到
            {
                ans = -1;
                break;
            }
        }
        if(tt != 1)
            ans = -1;
        if(ans == 0)
            printf("YES\n");
        if(ans == -1)
            printf("NO\n");
    }
    return 0;
}

ZOJ 3321 Circle(并查集啊)

标签：zoj   并查集   

原文地址：http://blog.csdn.net/u012860063/article/details/44876785