HDU 1689 (BFS+剪枝)

题目意思是检测最小奇数环 并输出结果

 

对于该类题目 分析了下复杂度 可以对各个点进行一一枚举 对于每次的枚举 记录访问到该点时的深度 当下一次访问时只要将当前深度加上访问到该点的深度就能计算出总共所用去的点数 使用2个队列  以形成逐层遍历（保证了路径的不重复） 感觉很不错 每次枚举计算出的最小值可以作为下一次的阀值 从而达到了缩小规模的目的

 

OK 下面是代码啦....

 

 

#include <iostream>
#include <queue>
using namespace std;
typedef struct 
{
    long next;
    long v;
}Edge;
long dp[1010];
long p[1010];
Edge e[40010];
long N,M;
bool vist[1010];
typedef struct  
{
    long v;
}Node;
queue<Node> q;
queue<Node> save;
inline void BFS()
{

    
    long i;
    long res=INT_MAX;

    for (i=1;i<=N;++i)
    {
        while (!q.empty())
        {
            q.pop();
        }

        memset(vist,0,sizeof(vist));
        
        Node st;
        st.v=i;
        dp[i]=0;
        q.push(st);
        vist[i]=true;

        long len=0;

        while (!q.empty()&&++len<res)
        {
            while (!save.empty())
            {
                save.pop();
            }

            while(!q.empty())//逐层扩展
            {
                Node t=q.front();
                q.pop();
                long j;
                for (j=p[t.v];j!=-1;j=e[j].next)
                {
                    long v=e[j].v;
                    if (vist[v])
                    {
                        long now=dp[v]+len;
                        if (now<res&&now&1)
                        {
                            res=now;
                        }
                    }
                    else
                    {
                        Node tt;
                        tt.v=v;
                        save.push(tt);
                        vist[v]=true;
                        dp[v]=len;
                    }
                }
            }

            while (!save.empty())
            {
                q.push(save.front());
                save.pop();
            }
        }
    }

    if (res==INT_MAX)
    {
        puts("Poor JYY.");
    }
    else
    {
        printf("JYY has to use %ld balls.\n",res);
    }
}

int main()
{
    long T;
    scanf("%ld",&T);
    long b=1;
    while (T--)
    {
        memset(p,-1,sizeof(p));        
        scanf("%ld %ld",&N,&M);
        long eid=0;
        while (M--)
        {
            long from,to;
            scanf("%ld %ld",&from,&to);
            
            e[eid].next=p[from];
            e[eid].v=to;
            p[from]=eid++;
            swap(from,to);
            e[eid].next=p[from];
            e[eid].v=to;
            p[from]=eid++;
            
        }
        printf("Case %ld: ",b);
        BFS();
        ++b;
    }
    return 0;
}