杭电1548 A strange lift（BFS过）（基础搜索）

A strange lift

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17299    Accepted Submission(s): 6496

Problem Description

There is a strange lift.The lift can stop can at every floor as you want, and there is a number Ki(0 <= Ki <= N) on every floor.The lift have just two buttons: up and down.When you at floor i,if you press the button "UP" , you will go up Ki floor,i.e,you will go to the i+Ki th floor,as the same, if you press the button "DOWN" , you will go down Ki floor,i.e,you will go to the i-Ki th floor. Of course, the lift can't go up high than N,and can't go down lower than 1. For example, there is a buliding with 5 floors, and k1 = 3, k2 = 3,k3 = 1,k4 = 2, k5 = 5.Begining from the 1 st floor,you can press the button "UP", and you'll go up to the 4 th floor,and if you press the button "DOWN", the lift can't do it, because it can't go down to the -2 th floor,as you know ,the -2 th floor isn't exist.
Here comes the problem: when you are on floor A,and you want to go to floor B,how many times at least he has to press the button "UP" or "DOWN"?

 

Input

The input consists of several test cases.,Each test case contains two lines.
The first line contains three integers N ,A,B( 1 <= N,A,B <= 200) which describe above,The second line consist N integers k1,k2,....kn.
A single 0 indicate the end of the input.

 

Output

For each case of the input output a interger, the least times you have to press the button when you on floor A,and you want to go to floor B.If you can't reach floor B,printf "-1".

 

Sample Input

5 1 53 3 1 2 50

 

Sample Output

3

搜索离不开搜的方式，我们知道，这个电梯无非两种走法：up or down。所以我们这里的走法是两种：

        for(int i=0;i<2;i++)        {            if(i==0)nex.x=now.x+f[now.x];//nex.x表示要走到的地方，now.x表示现在在的层数，f数组中存的是ki的值。这里表示up操作            if(i==1)nex.x=now.x-f[now.x];//这里表示down操作            if(nex.x>=1&&nex.x<=200&&vis[nex.x]==0)//范围题目中有所表明            {                vis[nex.x]=1;//走过的地方标记上                nex.output=now.output+1;//并且记录+1                s.push(nex);            }        }

然后是完整的AC代码：

 

#include<stdio.h>#include<queue>#include<string.h>using namespace std;int qidian,zongdian;int f[500];int vis[500];struct zuobiao{    int x,output;}now,nex;void bfs(int x){    queue<zuobiao >s;    memset(vis,0,sizeof(vis));    vis[x]=1;    now.x=x;    now.output=0;    s.push(now);    while(!s.empty())    {        now=s.front();        //printf("%d\n",now.x);        if(now.x==zongdian)        {            printf("%d\n",now.output);            return ;        }        s.pop();        for(int i=0;i<2;i++)        {            if(i==0)nex.x=now.x+f[now.x];            if(i==1)nex.x=now.x-f[now.x];            if(nex.x>=1&&nex.x<=200&&vis[nex.x]==0)            {                vis[nex.x]=1;                nex.output=now.output+1;                s.push(nex);            }        }    }    printf("-1\n");    return ;}int main(){    int n;    while(~scanf("%d",&n))    {        if(n==0)break;        scanf("%d%d",&qidian,&zongdian);        memset(f,0,sizeof(f));        for(int i=1;i<=n;i++)        {            scanf("%d",&f[i]);        }        bfs(qidian);    }}