POJ 3278: Catch That Cow
阿新 • • 發佈:2017-08-03
cat -s ostream body courier output 當前 ant 題意
Catch That Cow
(0 ≤ K ≤ 100,000) on the same number
line. Farmer John has two modes of transportation: walking and teleporting.
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 44613 | Accepted: 13946 |
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
* 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 KOutput
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
題意:有一個農民和一頭牛,他們在一個數軸上,牛在k位置保持不動,農戶開始時在n位置。設農戶當前在M位置,每次移動時有三種選擇:1.移動到M-1。2.移動到M+1位置;3.移動到M*2的位置。問最少移動多少次能夠移動到牛所在的位置。所以能夠用BFS來搜索這三個狀態,直到搜索到牛所在的位置。
#include<cstdio> #include<iostream> #include<cstring> #include<queue> #include<algorithm> using namespace std; const int N = 200100; int n, k; struct node { int x, step; }; queue<node> q; int vist[N]; void bfs() { int cow, ans; while(!q.empty()) { node tmp = q.front(); q.pop(); cow = tmp.x; ans = tmp.step; if(cow == k) { printf("%d\n",ans); return ; } if(cow >= 1 && !vist[cow - 1]) //要保證減1後有意義,所以要cow >= 1 減一的情況 { node temp; vist[cow - 1] = 1; temp.x = cow - 1; temp.step = ans + 1; q.push(temp); } if(cow <= k && !vist[cow + 1]) //加1的情況 { node temp; vist[cow + 1] = 1; temp.x = cow + 1; temp.step = ans + 1; q.push(temp); } if(cow <= k && !vist[cow * 2]) //乘二的情況 { node temp; vist[cow * 2] = 1; temp.x = 2 * cow; temp.step = ans + 1; q.push(temp); } } } int main() { while(~scanf("%d%d",&n,&k)) { while(!q.empty()) q.pop(); memset(vist,0,sizeof(vist)); vist[n] = 1; node t; t.x = n, t.step = 0; q.push(t); bfs(); } return 0; }
POJ 3278: Catch That Cow