poj 3278 bzoj 1646: [Usaco2007 Open]Catch That Cow 抓住那只牛（BFS）

1646: [Usaco2007 Open]Catch That Cow 抓住那只牛

Time Limit: 5 Sec Memory Limit: 64 MB
Submit: 1223 Solved: 583
[Submit][Status][Discuss]

Description

农夫约翰被通知，他的一只奶牛逃逸了！所以他决定，马上幽发，尽快把那只奶牛抓回来．

他们都站在数轴上．约翰在N(O≤N≤100000)处，奶牛在K(O≤K≤100000)处．约翰有

两种办法移动，步行和瞬移：步行每秒种可以让约翰从z处走到x+l或x-l处；而瞬移则可让他在1秒内从x处消失，在2x处出现．然而那只逃逸的奶牛，悲剧地没有发现自己的处境多么糟糕，正站在那儿一动不动．

那么，约翰需要多少时间抓住那只牛呢？

Input

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

仅有两个整数N和K.

Output

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

最短的时间．

Sample Input

5 17

Sample Output

搜索入门经典题

其实就是暴力

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
typedef struct
{int x;int time;
}state;
state now, temp;
queue<state> q;
int flag[150005];
int main(void)
{int n, m;while(scanf("%d%d", &n, &m)!=EOF){if(n>=m){printf("%d\n", n-m);continue;}memset(flag, 0, sizeof(flag));now.x = n, now.time = 0;q.push(now);flag[n] = 1;while(q.empty()==0){now = q.front();q.pop();if(now.x==m){while(q.empty()==0)q.pop();break;}if(now.x*2<=150000 && flag[now.x*2]==0){temp.x = now.x*2;temp.time = now.time+1;flag[temp.x] = 1;q.push(temp);}if(now.x-1>0 && flag[now.x-1]==0){temp.x = now.x-1;temp.time = now.time+1;flag[temp.x] = 1;q.push(temp);}if(now.x+1<=100000 && flag[now.x+1]==0){temp.x = now.x+1;temp.time = now.time+1;flag[temp.x] = 1;q.push(temp);}}printf("%d\n", now.time);}return 0;
}

poj 3278 bzoj 1646: [Usaco2007 Open]Catch That Cow 抓住那只牛（BFS）相关推荐

BZOJ 1646: [Usaco2007 Open]Catch That Cow
题目链接:http://www.lydsy.com/JudgeOnline/problem.php?id=1646 这道题我刚开始就直接硬生生地广搜,然后妥妥地TLE了,后来乖乖地加了一些剪枝... ...
Catch That Cow抓住那只奶牛（BFS，广度优先搜索）
# **Catch That Cow(抓住那只奶牛)** [题意]:农场主要去追一直逃跑的牛,每次移动有三种选择,位置加一,位置减一,位置乘二,要找到最快的方法让农夫追到牛(也就是移动的次数最少,这个 ...
Catch That Cow(抓住那头牛C++)
Catch That Cow Farmer John has been informed of the location of a fugitive cow and wants to catch he ...
poj 3268 bzoj 1631: [Usaco2007 Feb]Cow Party（最短路）
1631: [Usaco2007 Feb]Cow Party Time Limit: 5 Sec Memory Limit: 64 MB Submit: 855 Solved: 613 [Subm ...
Catch That Cow（农夫和牛）（BFS入门）（详解）
题目: 题目大意: 输入n和k表示农夫和牛的位置,n和k在区间[0,100000]间农夫有三种方式进行移动,每种方式需要的时间相同,都是一分钟 1.向前移动一米(+1) 2.向后移动一米(-1) 3 ...
C - Catch That Cow POJ - 3278
C - Catch That Cow POJ - 3278 首先是大暴搜+最优化剪枝(当当前搜索情况不如已有答案优时,返回) #include<cstdio> #include<al ...
POJ 3278 Catch That Cow
Catch That Cow Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 30924 Accepted: 9536 D ...
POJ 3278 Catch That Cow BFS
Catch That Cow Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 32071 Accepted: 9866 D ...
POJ 3278 Catch That Cow（BFS）
题目网址:http://poj.org/problem?id=3278 题目: Catch That Cow Time Limit: 2000MS Memory Limit: 65536K Tot ...