hdu 3478 //判聯通+二分圖判定

#include<stdio.h>
#include<string.h>
const int N = 100010;
struct NODE
{
    int v, next;
}edge[510000 * 2];
int clr[N], n, m;
int head[N], e;
void addedge(int u, int v)
{
    edge[e].v = v;
    edge[e].next = head[u];
    head[u] = e++;
}

//cnt判斷聯通性，flag=0表示能二分着色
int Clr_BFS(int s)
{
    int qu[N], f = -1, r = -1;
    int pos, tmp, cnt = 1, flag = 0;
    clr[s] = 1;
    qu[++r] = s;
    while(f != r)
    {
        pos = qu[++f];
        for(int p = head[pos]; p != -1; p = edge[p].next)
        {
            tmp = edge[p].v;
            if(clr[tmp] == clr[pos])
            {
                flag = 1; //return 0; //fail
                continue;
            }
            if(clr[tmp] == -1)
            {
                cnt++;
                clr[tmp] = clr[pos] ^ 1;
                qu[++r] = tmp;
            }
        }
    }
    if(cnt == n && flag)
        return 1;
    else return 0;
}

int main()
{
    int t, i, a, b, s, ca = 0;
    scanf("%d", &t);
    while(t--)
    {
        e = 0;
        memset(clr, -1, sizeof(clr));
        memset(head, -1, sizeof(head));
        scanf("%d%d%d", &n, &m, &s);

        for(i = 0; i < m; i++)
        {
            scanf("%d%d", &a, &b);
            addedge(a, b);
            addedge(b, a);
        }

        printf("Case %d: ", ++ca);
        if(Clr_BFS(s))
        {
            printf("YES\n");
        }
        else printf("NO\n");
    }
    return 0;
}