搜索------Catch That Cow

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从一个点a走到另一个点b，中间可以经过x+1，x-1,x*2，最少几步可以到达b点

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#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
#include<stack>
char str[105][105];
int dir[8][2] = {{1,1},{-1,1},{-1,-1},{1,-1}};
int n,k;
int a = 0;
using namespace std;
struct point
{
    int n,m;
};
int oo[111111];
int check(int n)
{
    if(n>=0&&n<=100000&&(!oo[n]))
        return 1;
    return 0;
}
int dfs(int x)
{
    queue<point>Q;
    point now,next;
    now.n = x;
    now.m = 0;
    oo[x] = 1;
    Q.push(now);
    while(!Q.empty())
    {
        now = Q.front();
        Q.pop();
        if(now.n == k)
        {
            return now.m;
        }
        next.n = now.n + 1;
        if(check(next.n))
        {
            oo[next.n] = 1;
            next.m = now.m+1;
            Q.push(next);
        }
        next.n = now.n - 1;
        if(check(next.n))
        {
            oo[next.n] = 1;
            next.m = now.m + 1;
            Q.push(next);
        }
        next.n = now.n * 2;
        if(check(next.n))
        {
            oo[next.n] = 1;
            next.m = now.m + 1;
            Q.push(next);
        }
    }
    return -1;
}
int main()
{
    while(~scanf("%d%d",&n,&k))
    {
        memset(oo,0,sizeof(oo));
        int ans = dfs(n);
        printf("%d\n",ans);
    }
    return 0;
}

搜索------Catch That Cow

标签：

原文地址：http://www.cnblogs.com/biu-biu-biu-/p/5683431.html