[kuangbin带你飞]专题一简单搜索C

最新推荐文章于 2018-08-27 10:21:52 发布

zjy2015302395

最新推荐文章于 2018-08-27 10:21:52 发布

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分类专栏： acm 带你飞系列基本算法

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本文介绍了一个趣味算法问题——如何通过行走与瞬移两种方式最快地追上静止不动的目标。通过广度优先搜索算法求解最短路径，并提供了一种高效的解决方案。

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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.
http://poj.org/problem?id=3278

题意：

三种走法，+1，-1，*2，

tip

Z = max(2*K,N);
如果目标在右侧不会走过目标2倍，因为往回走只能减1，浪费时间
如果目标在左侧不会超过起始点，不会再往右边走了哈

#include <cstdio>
#include <iostream>
#include <cstring>
#include <queue>
const int maxn = 200010;
using namespace std;
int N,K,Z;
bool vis[maxn];
int step[maxn];
void init(){
    memset(vis,false,sizeof(vis));
    Z = max(2*K,N);
}

void sov(){
    queue<int>q;
    q.push(N);
    step[N] = 0;
    vis[N] = true;
    while(!q.empty()){
        int tmp = q.front();
        q.pop();
        if(tmp == K){
            printf("%d\n",step[K]);
            return;
        }
        int x ;
        for(int i = 0 ; i < 3; i++){
            if(i == 0) x = tmp+1;
            if(i == 1) x = tmp-1;
            if(i == 2) x = tmp*2;
            if(x >=0 && x <= Z&&vis[x] == false){
                vis[x] = true;
                q.push(x);
                step[x] = step[tmp]+1;
            }
        }
    }
    return;
}
int main(){
    while(~scanf("%d%d",&N,&K)){
        init();
        sov();
    }
}