Catch That Cow POJ - 3278 [kuangbin带你飞]专题一 简单搜索
Posted 九月旧约
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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
Output
Sample Input
5 17
Sample Output
4
Hint
#include<iostream> #include<cstdio> #include<queue> #include<cstring> #include<cmath> using namespace std; #define maxn 100010 int n,k,vis[maxn]; struct node{ int x,step; }; void bfs(){ node p; p.x = n,p.step = 0; vis[n] = 1; queue<node> q; q.push(p); while(!q.empty()){ node tmp = q.front(); q.pop(); if(tmp.x == k){ cout << tmp.step << endl; return; } for(int i=0;i<3;i++){ int xx; if(i == 0){ xx = tmp.x + 1; } else if(i == 1){ xx = tmp.x - 1; } else xx = tmp.x * 2; if(xx < 0 || xx > maxn){ continue; } if(!vis[xx]){ vis[xx] = 1;//注意vis数组 node tp;//这里新建一个结构体对象,不要在原来的tmp上加减!!! 在这里调了一个半小时!!! tp.x = xx; tp.step = tmp.step + 1; q.push(tp); } } } } int main(){ while(cin >> n >> k){ memset(vis,0,sizeof(vis)); if(n<k){ bfs(); } else{ cout << n - k << endl; } } return 0; }
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