POJ - 3278
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 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
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 }
