poj3278（bfs）

 

题目链接：http://poj.org/problem?id=3278

分析：广搜，每次三种情况枚举一下，太水不多说了。

 

#include <cstdio>

#include <cstring>

#include <cmath>

#include <iostream>

#include <algorithm>

#include <queue>

#include <cstdlib>

#include <vector>

#include <set>

#include <map>

#define LL long long

#define mod 1000000007

#define inf 0x3f3f3f3f

#define N 100010

using namespace std;

struct node

{

    int x,step;

    bool operator<(const node a)const

    {

        return step>a.step;

    }

};

priority_queue<node>que;

int vis[N];

int judge(int x)

{

    return x>=0&&x<=100000;

}

node make_node(int a,int b)

{

    node temp;

    temp.x=a;temp.step=b;

    return temp;

}

int main()

{

    int n,k;

    while(scanf("%d%d",&n,&k)>0)

    {

        memset(vis,0,sizeof(vis));

        while(!que.empty())que.pop();

        que.push(make_node(n,0));

        vis[n]=1;

        int ans=0;

        while(!que.empty())

        {

            node now=que.top();

            que.pop();

            int x=now.x;//printf("%d %d\n",now.x,now.step);

            if(x==k)

            {

                ans=now.step;

                break;

            }

            if(judge(x+1)&&!vis[x+1])

            {

                vis[x+1]=1;que.push(make_node(x+1,now.step+1));

            }

            if(judge(x-1)&&!vis[x-1])

            {

                vis[x-1]=1;que.push(make_node(x-1,now.step+1));

            }

            if(judge(2*x)&&!vis[x*2])

            {

                vis[x*2]=1;que.push(make_node(2*x,now.step+1));

            }

        }

        printf("%d\n",ans);

    }

}

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