POJ 3278 Catch That Cow

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给N给K 有三种操作 N+1 N-1 N*2 让N==K 输出步骤数

宽搜。

#include <iostream>#include <cstdio>#include <cstring>using namespace std;int n,k;struct node{    int deep;  //记录步骤数    int w;    //记录数值}queue[1000005];int vis[1000005];int main(){    while(scanf("%d%d",&n,&k)!=EOF)    {        int head=0,tail=0;        memset(vis,0,sizeof(vis));        vis[n]=1;        queue[tail].deep=0;        queue[tail++].w=n;        while(head < tail)        {            node t = queue[head++];            if(t.w == k)            {                printf("%d\n",t.deep);                break;            }            if(t.w-1 >= 0 && !vis[t.w-1])            {                queue[tail].w = t.w-1;                queue[tail++].deep = t.deep+1;                vis[t.w-1]=1;            }            if(t.w <= k && !vis[t.w*2])            {                queue[tail].w = t.w*2;                queue[tail++].deep = t.deep+1;                vis[t.w*2]=1;            }            if(t.w <=k  && !vis[t.w+1])            {                queue[tail].w = t.w+1;                queue[tail++].deep = t.deep+1;                vis[t.w+1]=1;            }        }    }    return 0;}