poj3278: Catch That Cow结题报告

Catch That Cow Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 96169   Accepted: 30168 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

代码：

#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;
bool visit[200002];
typedef pair<int, int> PAIR;
int bfs(int n, int k){
queue<PAIR> que;
que.push(PAIR(n, 0));
if(n == k) return 0;
visit
= true;
while(!que.empty()){
PAIR p = que.front();
que.pop();
if(p.first == k) return p.second;
if(p.first - 1 >= 0 &&!visit[p.first - 1]){
que.push(PAIR(p.first - 1, p.second + 1));
visit[p.first - 1] = true;
}
if(p.first + 1 <= 100000 && !visit[p.first + 1]){
que.push(PAIR(p.first + 1, p.second + 1));
visit[p.first + 1] = true;
}
if((p.first<<1) <= 100000 && !visit[p.first<<1]){
que.push(PAIR(p.first<<1, p.second + 1));
visit[p.first<<1] = true;
}
}
return -1;
}
int main(){
int n, k;
scanf("%d%d",&n,&k);
memset(visit, 0, sizeof(visit));
int ans = bfs(n, k);
printf("%d\n",ans);
return 0;
}