poj 3278 暴力BFS

最新推荐文章于 2019-03-28 22:54:37 发布

原创最新推荐文章于 2019-03-28 22:54:37 发布 · 336 阅读

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本文介绍了一道经典的算法题目“CatchThatCow”的解题思路。该题要求在一个数轴上，通过移动和瞬移操作从起点N到达目标点K，并求出所需的最短时间。文章详细介绍了使用广度优先搜索(BFS)算法来解决这一问题的方法。

Catch That Cow
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 63977 Accepted: 20085
Description

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 100,000) on a number line and the cow is at a point K (0 ≤ K ≤ 100,000) on the same number line. Farmer John has two modes of transportation: walking and teleporting.

Walking: FJ can move from any point X to the points X - 1 or X + 1 in a single minute
Teleporting: FJ can move from any point X to the point 2 × X in a single minute.

If the cow, unaware of its pursuit, does not move at all, how long does it take for Farmer John to retrieve it?

Input

Line 1: Two space-separated integers: N and K
Output

Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.
Sample Input

5 17
Sample Output

4
Hint

The fastest way for Farmer John to reach the fugitive cow is to move along the following path: 5-10-9-18-17, which takes 4 minutes.

题意：
在一个数轴上，给两个点N，K，问N最少经过几次操作可以到K（操作可以是向后走一步，向前走一步，或向前跳淘到2x处，x为当前位置，问最少几次可以到K）
解法：
直接BFS

 <iostream>
#include <algorithm>
#include <cstdio>
#include <queue>
#include <cstring>

using namespace std;

typedef struct{
    int v,h;
}Po;

//queue<Po>qu;
int n,k;
int p[200010];

int main()
{
//    freopen("data.out","w",stdout);
    while(scanf("%d%d",&n,&k)!=EOF){
        queue<Po>qu;
        memset(p,0,sizeof(p));
        Po S;
        S.h = 0;
        S.v = n;
        qu.push(S);
        Po te;
        while(!qu.empty()){
            te = qu.front();qu.pop();
            if(te.v == k){
                    break;
                }
            if(0<=te.v + 1&&te.v+1<=100000){
                Po ter = te;
                ter.v++;
                ter.h++;
                if(!p[ter.v]){
                    qu.push(ter);
                    p[ter.v] = 1;
                }

            }
            if(0<=te.v - 1&&te.v - 1<=100000){
                Po ter = te;
                ter.v--;
                ter.h++;
                if(!p[ter.v]){
                    qu.push(ter);
                    p[ter.v] = 1;
                }
            }
            if(0<=te.v*2&&te.v*2<=100000){
                Po ter = te;
                ter.v*=2;
                ter.h++;
                if(!p[ter.v]){
                    qu.push(ter);
                    p[ter.v] = 1;
                }
            }
        }
        printf("%d\n",te.h);
    }
    return 0;
}