POJ - 3278 - Catch That Cow （BFS）

题目传送：Catch That Cow

思路：BFS找最小步数，用一个结构体存下当前结点的数值以及当前步数

AC代码：

#include <cstdio>
#include <cstring>
#include <string>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <set>
#include <deque>
#include <cctype>
#define LL long long
#define INF 0x7fffffff
using namespace std;

int n, k;

bool vis[200005];

struct node {
	int num;
	int cnt;
	node(int n, int c) : num(n), cnt(c) {};
};

int bfs(int n, int k) {
	if(n == k) return 0;
	memset(vis, false, sizeof(vis));
	queue<node> que;
	que.push(node(n, 0));
	vis[n] = true;
	while(!que.empty()) {
		node t = que.front();
		que.pop();
		node n1 = node(t.num - 1, t.cnt + 1);
		node n2 = node(t.num + 1, t.cnt + 1);
		node n3 = node(t.num * 2, t.cnt + 1);
		if(n1.num == k) return n1.cnt;
		if(n2.num == k) return n2.cnt;
		if(n3.num == k) return n3.cnt;
		if(n1.num >= 0 && n1.num <= 150005 && !vis[n1.num]) {
			//重要剪枝，num小于0的时候肯定不是正解，还有num大于150000时肯定也不是正解
			que.push(n1);
			vis[n1.num] = true;//入队列的时候就要赋值为已经访问过，不然会超时！
		}
		if(n2.num >= 0 && n2.num <= 150005 && !vis[n2.num]) {
			que.push(n2);
			vis[n2.num] = true;
		}
		if(n3.num >= 0 && n3.num <= 150005 && !vis[n3.num]) {
			que.push(n3);
			vis[n3.num] = true;
		}
	}
}

int main() {
	while(scanf("%d %d", &n, &k) != EOF) {
		printf("%d\n", bfs(n, k));
	}
	return 0;
}