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POJ - 3278

时间:2014-07-22 22:47:37      阅读:191      评论:0      收藏:0      [点我收藏+]

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Time Limit: 2000MS  Memory Limit: 65536KB  64bit IO Format: %I64d & %I64u

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 - 1 or + 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

双向宽搜,判重数组开到20万。
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <algorithm>
 6 #include <string>
 7 #include <vector>
 8 #include <queue>
 9 #include <stack>
10 #include <set>
11 #include <map>
12 #include <limits.h>
13 #include <bitset>
14 #define F "%I64d"
15 using namespace std;
16 typedef long long ll;
17 int dp[2][200000],N,K;
18 queue<int> QN,QK;
19 int bfs()
20 {
21     while(!QN.empty())QN.pop();
22     while(!QK.empty())QK.pop();
23     memset(dp,-1,sizeof dp);
24     QN.push(N);
25     QK.push(K);
26     dp[0][N] = dp[1][K] = 0;
27     while(1)
28     {
29         if(QN.front() > 1 && dp[0][QN.front() - 1] == -1)
30         {
31             dp[0][QN.front() - 1] = dp[0][QN.front()] + 1;
32             if(dp[1][QN.front() - 1] != -1)return dp[0][QN.front() - 1] + dp[1][QN.front() - 1];
33             QN.push(QN.front() - 1);
34         }
35         if(QN.front() < 100000 && dp[0][QN.front() + 1] == -1)
36         {
37             dp[0][QN.front() + 1] = dp[0][QN.front()] + 1;
38             if(dp[1][QN.front() + 1] != -1)return dp[0][QN.front() + 1] + dp[1][QN.front() + 1];
39             QN.push(QN.front() + 1);
40         }
41         if(QN.front() < 100000 && dp[0][QN.front() << 1] == -1)
42         {
43             dp[0][QN.front() << 1] = dp[0][QN.front()] + 1;
44             if(dp[1][QN.front() << 1] != -1)return dp[0][QN.front() << 1] + dp[1][QN.front() << 1];
45             QN.push(QN.front() << 1);
46         }
47         if(QK.front() > 1 && dp[1][QK.front() - 1] == -1)
48         {
49             dp[1][QK.front() - 1] = dp[1][QK.front()] + 1;
50             if(dp[0][QK.front() - 1] != -1)return dp[1][QK.front() - 1] + dp[0][QK.front() - 1];
51             QK.push(QK.front() - 1);
52         }
53         if(QK.front() < 100000 && dp[1][QK.front() + 1] == -1)
54         {
55             dp[1][QK.front() + 1] = dp[1][QK.front()] + 1;
56             if(dp[0][QK.front() + 1] != -1)return dp[1][QK.front() + 1] + dp[0][QK.front() + 1];
57             QK.push(QK.front() + 1);
58         }
59         if((QK.front() & 1) == 0 && dp[1][QK.front() >> 1] == -1)
60         {
61             dp[1][QK.front() >> 1] = dp[1][QK.front()] + 1;
62             if(dp[0][QK.front() >> 1] != -1)return dp[1][QK.front() >> 1] + dp[0][QK.front() >> 1];
63             QK.push(QK.front() >> 1);
64         }
65         QN.pop();
66         QK.pop();
67     }
68 }
69 int main()
70 {
71 N:    while(~scanf("%d%d",&N,&K))
72     {
73         if(abs(N - K) <= 1){printf("%d\n",abs(N - K));continue;}
74         if((N << 1) == K){puts("1");continue;}
75         if(K < N){printf("%d\n",N - K);continue;}
76         printf("%d\n",bfs());
77     }
78     return 0;
79 }

POJ - 3278,布布扣,bubuko.com

POJ - 3278

标签:des   style   blog   http   color   os   

原文地址:http://www.cnblogs.com/gangduo-shangjinlieren/p/3859588.html

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