HDU1548 A strange life BFS

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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 5

3 3 1 2 5

Sample Output

3

題目大意：一個奇怪的電梯，就是你要從A坐到B，每層電梯都有一個按鈕,按鈕上有數字，數字表示可以向上或者向下下降幾層.k[i]表示在第i層的數字是k[i]. 問最少按幾次按鈕。

分析：這種題目吧，一看最少的話，簡單題目就是BFS。 每一步有兩中狀态，分别進隊，直到找到答案為止，注意在選擇向上還是向下的時候 一定要判斷向上的是否超過n，向下是否比1小。還有就是巧妙的利用vis[]數組，既然标記了,也起到了儲存答案的作用

#include<iostream>
#include<queue>
#include<cstring>
#include<cstdio>
using namespace std;

const int maxn=;

int vis[maxn];
queue <int> q;
int a,b,n;
int k[];

void bfs(int x){
    q.push(x);
    while (!q.empty()) {
        int xx =q.front();
        q.pop();
        int down = xx-k[xx];
        int up = xx+k[xx];
        if(up<=n&&!vis[up]){
            vis[up]=vis[xx]+;
            q.push(up);
        }
        if(down>=&&!vis[down]){
            vis[down] = vis[xx]+;
            q.push(down);
        }
        if(up==b||down == b) break;
    }
}

int main()
{
    while(scanf("%d",&n),n){
        memset(vis,,sizeof vis);
        memset(k,,sizeof k);
        scanf("%d %d",&a,&b);
        while(!q.empty()){
            q.pop();
        }
        vis[a] = ;
        for(int i =  ;i <= n;i++){
            scanf("%d",&k[i]);
        }
        bfs(a);
        cout << vis[b]- << endl;
    }
    return ;
}