Catch That Cow
Time Limit:2000MSMemory Limit:65536K
Total Submissions:52335Accepted:16416
Description
Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a pointN(0 ≤N≤ 100,000) on a number line and the cow is at a pointK(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 pointXto the pointsX- 1 orX+ 1 in a single minute* Teleporting: FJ can move from any pointXto the point 2 ×Xin 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:NandK
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.
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;}*/
大多数人想要改造这个世界，但却罕有人想改造自己。
