POJ 3287 Catch That Cow(bfs)

看懂意思就很简单了，给你一个数，每次只能加一或者减一或者乘2，问最少几次才能变成另一个数，就是一个简单的bfs过程，看代码吧。

AC代码：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
using namespace std;
const int MAXN = 100005;
struct Node{int x,step;
}Now,Next,S,E;
bool vis[MAXN];                         // 用于标记这个数是否出现过int bfs(){queue<Node> q;memset(vis,0,sizeof(vis));S.step = 0;q.push(S);while(!q.empty()){Now = q.front();q.pop();if(Now.x == E.x){                   // 结束的时候返回步数return Now.step;}for(int i=0;i<3;i++){               // 模拟变换过程if(i==0){Next.x = Now.x + 1;}else if(i==1){Next.x = Now.x - 1;}else{Next.x = Now.x * 2;}if(Next.x >= 0 && Next.x <= 100000 && vis[Next.x] == 0){vis[Next.x] = 1;Next.step = Now.step + 1;q.push(Next);}}}return -1;
}int main()
{while(~scanf("%d%d",&S.x,&E.x)){int temp = bfs();printf("%d\n",temp);}return 0;
}
/***[来源] POJ 3278[题目] Catch That Cow[大意]给你两个数n,m，n每次只能加一或者减一或者乘以2，问最少需要多少次才能变成m[输入]5 17[输出]4
*/

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