POJ - 3278 Catch That Cow （BFS）

题目大意：FJ 初始在点 N，奶牛初始在点 K，两点在同一条线上。

FJ 有两种移动方式：

1、x+1 或 x-1

2、2*x

奶牛不动，问最少移动几次 FJ 可以抓住奶牛。

解题思路：三个入口的 BFS，注意一下边界，不控制一下会 RE

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<cmath>
#include<string.h>
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
const int INF = 0x3f3f3f3f;
const int NINF = -INF -1;
const int MAXN = 100000+10;
using namespace std;
int N, K, tmp;
struct point {
int x;
int t;
point(int a = 0, int b = 0):x(a),t(b){}
};
point p[MAXN];
int vis[MAXN];
int bfs(int x) {
if (x == K) return 0;
p[0] = point(x, 0);

int t1 = 0, t2 = 1;
while (t1 < t2) {
point now = p[t1++];
//      printf("%d\n", now.x);
for (int i = 0; i < 3; i++) {
if (i == 0) tmp = now.x-1;
else if (i == 1) tmp = now.x+1;
else if (i == 2) tmp = now.x*2;
if (tmp < 0 || tmp > MAXN) continue;
if (!vis[tmp]) {
if (tmp == K) return now.t+1;
p[t2++] = point(tmp, now.t+1);
vis[tmp] = 1;
}
}
}
return -1;
}
int main() {
while (scanf("%d%d", &N, &K) != EOF) {
memset(vis, 0, sizeof(vis));
printf("%d\n", bfs(N));
}
return 0;
}