本文主要是介绍poj 3278 Catch That Cow 【bfs】，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 52335 | Accepted: 16416 |
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
Output
Sample Input
5 17
Sample Output
4
Hint
简单bfs， 每个状态可以转换成其他三个状态枚举就好了。
原来是打算用spfa， 但是TL了
代码：
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#include <vector>
const int M = 1e5+50;
using namespace std;struct node{int x;int step;bool operator < (const node &a) const {return step > a.step;}
};
bool vis[M];int bfs(int n, int k){memset(vis, 0, sizeof(vis));if(n == k) return 0;node st;st.x = n; st.step = 0;priority_queue<node> q;q.push(st);while(!q.empty()){node temp = q.top();q.pop();node cur;cur.step = temp.step+1;cur.x = temp.x+1;if(cur.x == k) return cur.step;if(cur.x < M&&!vis[cur.x]){vis[cur.x] = 1;q.push(cur);}cur.x = temp.x-1;if(cur.x == k) return cur.step;if(cur.x > 0&&cur.x < M&&!vis[cur.x]){vis[cur.x] = 1;q.push(cur);}cur.x = temp.x*2;if(cur.x == k) return cur.step;if(cur.x < M&&!vis[cur.x]){vis[cur.x] = 1;q.push(cur);}}
}int main(){int n, k;while(scanf("%d%d", &n, &k) == 2){printf("%d\n", bfs(n, k));}
}
/*vector <int >map[M*4];
int d[M];
bool vis[M];/*int bfs(int n, int k){memset(vis, 0, sizeof(vis));priority_queue<node >q;node st;st.x = n; st.step = 0;vis[n] = 1;q.push(st);if(n == k) return 0;while(!q.empty()){node temp = q.top();q.pop();node cur;cur.step = temp.step+1;cur.x = temp.x+1;if(cur.x == k) return cur.step;if(cur.x > 0&&cur.x < M){vis[cur.x] = 1; q.push(cur);}cur.x = temp.x-1;if(cur.x == k) return cur.step;if(cur.x > 0&&cur.x < M){vis[cur.x] = 1; q.push(cur);}cur.x = temp.x*2;if(cur.x == k) return cur.step;if(cur.x > 0&&cur.x < M){vis[cur.x] = 1; q.push(cur);}}
}*//*void spfa(int n, int k){memset(vis, 0, sizeof(vis));memset(d, 'a', sizeof(d));queue<int >q;q.push(n);d[n] = 0;vis[n] = 1;while(!q.empty()){int temp = q.front();q.pop();for(int i = 0; i < map[temp].size(); ++ i){if(d[map[temp][i]] > d[temp]+1){d[map[temp][i]] = d[temp]+1;if(!vis[map[temp][i]]){vis[map[temp][i]] = 1;q.push(map[temp][i]);} }}}
}int main(){int n, k;while(scanf("%d%d", &n, &k) == 2){for(int i = 1; i < M; ++ i){map[i].clear();if(i-1 > 0) map[i].push_back(i-1);if(i+1 < M) map[i].push_back(i+1);if(i*2 < M) map[i].push_back(i*2);}spfa(n, k);//for(int i = 0; i < M; ++ i) vis[i] = 0, map[i].clear();//for()printf("%d\n", d[k]);}return 0;
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