M - Catch That Cow POJ - 3278

Think:

1BFS

2反思:注意数组越界+注意标记走过的点

3C++中的STL的queue练习

M - Catch That Cow POJ - 3278

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.

以下为Runtime Error代码——数组越界

#include <cstdio>
#include <cstring>
#include <queue>

using namespace std;

struct node{
int x;
int step;
}a, b;

int v[240000];

void BFS(int n, int m);

int main(){
int n, m;
while(scanf("%d %d", &n, &m) != EOF){
BFS(n, m);
}
return 0;
}
void BFS(int n, int m){
queue<struct node> q;
a.x = n;
a.step = 0;
q.push(a);
v
= 1;
while(!q.empty()){
a = q.front();
q.pop();
if(a.x == m){
printf("%d\n", a.step);
break;
}
if(a.x < m && v[a.x+1] == 0){
b.x = a.x + 1;
b.step = a.step + 1;
q.push(b);
v[b.x] = 1;
}
if(a.x >= 1 && v[a.x-1] == 0){
b.x = a.x - 1;
b.step = a.step + 1;
q.push(b);
v[b.x] = 1;
}
if(v[a.x*2] == 0){
b.x = a.x*2;
b.step = a.step + 1;
q.push(b);
v[b.x] = 1;
}
}
}

以下为Accepted代码

#include <cstdio>
#include <cstring>
#include <queue>

using namespace std;

struct node{
int x;
int step;
}a, b;

int v[240000];

void BFS(int n, int m);

int main(){
int n, m;
while(scanf("%d %d", &n, &m) != EOF){
BFS(n, m);
}
return 0;
}
void BFS(int n, int m){
queue<struct node> q;
a.x = n;
a.step = 0;
q.push(a);
v
= 1;
while(!q.empty()){
a = q.front();
q.pop();
if(a.x == m){
printf("%d\n", a.step);
break;
}
if(a.x < m && v[a.x+1] == 0){
b.x = a.x + 1;
b.step = a.step + 1;
q.push(b);
v[b.x] = 1;
}
if(a.x >= 1 && v[a.x-1] == 0){
b.x = a.x - 1;
b.step = a.step + 1;
q.push(b);
v[b.x] = 1;
}
if(a.x <= m && v[a.x*2] == 0){
b.x = a.x*2;
b.step = a.step + 1;
q.push(b);
v[b.x] = 1;
}
}
}