Catch That Cow
Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 22964 Accepted Submission(s): 6568
Problem 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.
题目大意是一行输入n 和 k 代表农夫和牛的位置，农夫可以向前或向后走一步，或者到2*他现在位置的那个位置，问最少时间（几步）能抓到那头牛。
总结：
- 要多组输入！但题目没说。
- bfs一开始一定要 memset 那个标记数组啊！ 不然就wa啊！！
下面附上ac代码：
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <string.h>
#include <ctype.h>
#include <queue>
#include <cmath>
#define INF 0x3f3f3f3f
#define mod 1000000007
using namespace std;
typedef long long int ll;
int sgn[210000];
int n,k;
struct node {
int x,step;
};
bool check(int x) {
if(x < 0 || x > 100000) {return false;}
if(sgn[x] == 1) {return false;}
return true;
}
int bfs() {
memset(sgn,0,sizeof(sgn));
node cur,next;
cur.x = n;
sgn[cur.x] = 1;
cur.step = 0;
queue<node>q;
q.push(cur);
while(!q.empty()) {
cur = q.front();
q.pop();
if(cur.x == k) {
return cur.step;
}
next.x = cur.x + 1;
if(check(next.x)) {
next.step = cur.step + 1;
sgn[next.x] = 1;
q.push(next);
}
next.x = cur.x - 1;
if(check(next.x)) {
next.step = cur.step + 1;
sgn[next.x] = 1;
q.push(next);
}
next.x = cur.x * 2;
if(check(next.x)) {
next.step = cur.step + 1;
sgn[next.x] = 1;
q.push(next);
}
}
}
int main() {
while(cin >> n >> k) {
cout << bfs() << endl;
}
return 0;
}