HDU 1548 A strange lift (BFS)

题目链接：HDU 1548

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 5
3 3 1 2 5
0

Sample Output

3

Solution

题意

一幢楼有 n 层，电梯在每一层都可以上或者下 k 层，问能不能从 A 层到达 B 层，如果能输出最少坐几次电梯，如果不能输出 -1。

题解

BFS 水题

使用 BFS 求最少次数，每次搜索两种状态：上或下，如果能到达 B 层，就输出最先到达的步数，如果不能输出-1。

注：此题还可用 Dijkstra 求解。

Code

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll inf = 0x3f3f3f3f;
const int maxn = 1010;

int n, A, B;
int k[maxn], dis[maxn], vis[maxn];

int bfs(int s) {
    queue<int> que;
    que.push(s);
    while(!que.empty()) {
        int x = que.front();
        que.pop();
        if(x == B) return dis[x];  // 如果到达 B 层就输出
        for(int i = -1; i <= 1; i += 2) {
            int tmp = x + i * k[x];  // 往上或者往下
            // 判断楼层是否合法
            if(tmp > 0 && tmp <= n && !vis[tmp]) {
                dis[tmp] = dis[x] + 1;
                vis[tmp] = 1;
                que.push(tmp);
            }
        }
    }
    return -1;  // 无法到达 B 层
}

int main() {
    while(~scanf("%d",&n) && n) {
        // 输入
        scanf("%d%d", &A, &B);
        for(int i = 1; i <= n; ++i) {
            scanf("%d", &k[i]);
        }
        // 初始化
        for(int i = 1; i <= n; ++i) {
            dis[i] = inf;
            vis[i] = 0;
        }
        dis[A] = 0;
        vis[A] = 1;
        // 搜索并输出
        printf("%d
", bfs(A));
    }
    return 0;
}