HDU1548 Building Roads

A strange lift

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

第一种解法：使用BFS，这里需要考虑到队列中楼层重复的问题，所有设置了一个vis来避免相同数据加入。

#include <iostream>
#include<vector>
#include<bits/stdc++.h>
#include<queue>
using namespace std;
bool vis[210];
struct node{
    int num;
    int step;
    node (){};
    node(int num,int step){
        this->step= step;
        this->num=num;
    }
};
void bfs(int n,int a,int b,node* floor){
    int flag=0;
    memset(vis,0,sizeof(vis));
    queue<node>que;
    node st(a,0) ;
    que.push(st);
    while (!que.empty()){
        node start = que.front();
        que.pop();
        vis[start.num]=1;
        if(start.num==b) {
            cout<<start.step<<endl;
            return;
        }
        for (int i = 0; i < 2; i++){
            if(i==0){
                int num = start.num-floor[start.num].num;
                if(num>=1&&num<=n&&!vis[num]) {
                    que.push( node(num,start.step+1));
                }
            }
            else{
                int num = start.num+floor[start.num].num;
                if(num>=1&&num<=n&&!vis[num]) {
                    que.push(node(num,start.step+1));
                }
            }
        }
    }
    if(!flag)cout<<"-1"<<endl;
}

int main(){
    int n,a,b,k;
    while (cin>>n,n>0){
        cin>>a>>b;
        node floor[210];
        for (int i = 1; i <= n; i++){
            cin>>k;
            floor[i].num=k;
            floor[i].step=0;
        }
        bfs(n,a,b,floor);
    }
}

第二种解法:使用最短路dijkstra

#include <stdio.h>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <queue>
using namespace std;
const int N =205;
const int INF = 9999999;
int n;
int graph[N][N];
int dist[N];
bool vis[N];
void dijkstra(int s){
    memset(vis,false,sizeof(vis));
    for(int i=1;i<=n;i++){
        dist[i] = graph[s][i];
    }
    for(int i=1;i<=n;i++){
        int mindis = INF;
        int mark;
        for(int j=1;j<=n;j++){
            if(!vis[j]&&dist[j]<mindis){
                mark = j;
                mindis = dist[j];
            }
        }
        vis[mark] = true;
        for(int j=1;j<=n;j++){
            if(!vis[j]&&dist[j]>dist[mark]+graph[mark][j]){
                dist[j] = dist[mark]+graph[mark][j];
            }
        }
    }
}
int main(){
    while(scanf("%d",&n)!=EOF,n){
        int s,t;
        scanf("%d%d",&s,&t);
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                if(i==j) graph[i][j]=0;
                else graph[i][j] = INF;
            }
        }
        for(int i=1;i<=n;i++){
            int num;
            scanf("%d",&num);
            if(i-num>=1) graph[i][i-num] = 1;
            if(i+num<=n) graph[i][i+num] = 1;
        }
        dijkstra(s);
        if(dist[t]>=INF) printf("-1\n");
        else printf("%d\n",dist[t]);
    }
    return 0;
}