POJ.3278 Catch That Cow (BFS)

POJ.3278 Catch That Cow (BFS)

题意分析

求最少的操作次数，暴力就用BFS。
这题坑点挺多，一开始交上去无限RE，摸不着头脑，后来发现后数组越界了。因为存在-1的操作和*2的操作 ，不加判断就直接越界。
这告诉我们一个道理，光把数组开大点，是没有用的。

代码总览

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <sstream>#include <set>#include <map>#include <queue>#include <stack>#include <cmath>#define INF 0x3f3f3f3f#define nmax 400010#define MEM(x) memset(x,0,sizeof(x))using namespace std;int num, tar;int ans;struct s{    int num;    int time;};bool visit[nmax];void bfs(int num,int time){    queue<s> q;    s now = {num,time},temp;    visit[num] = true;    q.push(now);    while(!q.empty()){        temp = q.front();q.pop();        if(temp.num == tar){            ans = temp.time;            break;        }        if(!visit[temp.num+1]){            now.num = temp.num+1;            now.time = temp.time+1;            visit[temp.num+1] = true;            q.push(now);        }        if(temp.num-1>=0 && !visit[temp.num-1]  ){            now.num = temp.num-1;            now.time = temp.time+1;            visit[temp.num-1] = true;            q.push(now);        }        if(temp.num*2<nmax && !visit[temp.num*2] &&temp.num*2 >=0 ){            now.num = temp.num*2;            now.time = temp.time+1;            visit[temp.num*2] = true;            q.push(now);        }    }}int main(){    //freopen("in.txt","r",stdin);    while(scanf("%d %d",&num,&tar)!=EOF){       MEM(visit);       ans = INF;       bfs(num,0);       printf("%d\n",ans);    }    return 0;}