POJ3278 BFS

题目内容 
Catch That Cow
Time Limit: 2000MS        Memory Limit: 65536K
Total Submissions: 26330        Accepted: 8122

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.

Source
USACO 2007 Open Silver

 

很纠结的题，算法没有错误却不断WA，最后改成C++编译器AC。错误应该出现在STL的queue里，具体不知。。。手动模拟队列AC。

代码如下（只能用C++编译）：

View Code 
  1 #include<cstdio>
  2 
  3 #include<cstring>
  4 
  5 #include<queue>
  6 
  7 #define MAXN 100001
  8 
  9 using namespace std;
 10 
 11 long n,k;
 12 
 13 bool visited[MAXN];
 14 
 15 long step[MAXN];
 16 
 17 long bfs(long start)
 18 
 19 {
 20 
 21     queue<long>q;
 22 
 23     q.push(start);
 24 
 25     visited[start]=true;
 26 
 27     long ns,t;
 28 
 29     while(!q.empty())
 30 
 31     {
 32 
 33         ns=q.front();
 34 
 35         q.pop();
 36 
 37         if(ns==k)
 38 
 39             return step[ns];
 40 
 41         t=ns+1;
 42 
 43         if(t>=0&&t<=100000&&visited[t]==false)
 44 
 45         {
 46 
 47             step[t]=step[ns]+1;
 48 
 49             if(t==k)
 50 
 51                 return step[t];
 52 
 53             visited[t]=true;
 54 
 55             q.push(t);
 56 
 57         }
 58 
 59         t=ns-1;
 60 
 61         if(t>=0&&t<=100000&&visited[t]==false)
 62 
 63         {
 64 
 65             step[t]=step[ns]+1;
 66 
 67             if(t==k)
 68 
 69                 return step[t];
 70 
 71             visited[t]=true;
 72 
 73             q.push(t);
 74 
 75         }
 76 
 77         t=ns*2;
 78 
 79         if(t>=0&&t<=100000&&visited[t]==false)
 80 
 81         {
 82 
 83             step[t]=step[ns]+1;
 84 
 85             if(t==k)
 86 
 87                 return step[t];
 88 
 89             visited[t]=true;
 90 
 91             q.push(t);
 92 
 93         }
 94 
 95     }
 96 
 97 }
 98 
 99 void init()
100 
101 {
102 
103     memset(visited,false,sizeof(visited));
104 
105     memset(step,0,sizeof(step));
106 
107 }
108 
109 int main(void)
110 
111 {
112 
113     while(scanf("%ld%ld",&n,&k)==2)
114 
115     {
116 
117         init();
118 
119         printf("%ld\n",bfs(n));
120 
121     }
122 
123     return 0;
124 
125 }



posted on 2011-10-30 15:41  SilVeRyELF  阅读(246)  评论(0)    收藏  举报

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