HDU 1548 A strange lift（Dijkstra、BFS、DP）
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"?
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.
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
3http://acm.hdu.edu.cn/showproblem.php?pid=1548
可以用到的算法：Dijkstra、BFS、DP
题意:
有一个特别的电梯，第i层有一个对应的数字ki， 对于第i层按上升键up可升上到i+k[i]层，按下降键down到达i-k[i]层，到达的楼层最高不能超过n层，最低不能小于1层。给你一个起点A和终点B，问最少要按几次上升键或者下降键到达目的地。
思路一：
最短路：把每一层都看成一个节点，问题就可以变成求起点到终点的最短路径问题。
用Dijkstra算法和BFS算法都可以解。Dijkstra： 100+MS
#include<iostream> using namespace std; #define INF 999999 int map[500][500]; int dist[500],used[500]; int n; void Dijkstra(int v) { int i,j,pos,min; memset(used,0,sizeof(used)); for(i=1;i<=n;i++) dist[i]=map[v][i]; used[v]=1; dist[v]=0;//floor A==floor B(起点就是终点)的情况 for(j=1;j<n;j++) { pos=0; min=100000000; for(i=1;i<=n;i++) if(dist[i]<min&&used[i]==0) { pos=i; min=dist[i]; } used[pos]=1; for(i=1;i<=n;i++) if(dist[i]>dist[pos]+map[pos][i]) dist[i]=dist[pos]+map[pos][i]; } } int main() { int s,t,i,j; int a[500]; while(cin>>n,n) { cin>>s>>t; memset(map,INF,sizeof(map)); for(i=1;i<=n;i++) { cin>>a[i]; if(i+a[i]<=n) map[i][i+a[i]]=1;//构图必须是单向边 if(i-a[i]>=1) map[i][i-a[i]]=1; } Dijkstra(s); if(dist[t]>INF)//If you can't reach floor B,printf "-1" cout<<-1<<endl; else cout<<dist[t]<<endl; } return 0; }
BFS：15MS#include<cstdio> #include<queue> #include<cstring> using namespace std; #define maxn 250 bool visit[maxn]; int floor[maxn][2];//floor[i][0]表示第i层向上能到的楼层，floor[i][1]则表示向下能到的楼层 int n,start,end; struct node { int pos,t; }temp,p; queue<node> q; int BFS() { memset(visit,false,sizeof(visit)); while(!q.empty()) { temp=q.front(); q.pop(); visit[temp.pos]=true; if(temp.pos==end) return temp.t; int up=floor[temp.pos][0],down=floor[temp.pos][1]; if(up!=-1&&!visit[up]) { p.pos=up; p.t=temp.t+1; q.push(p); } if(down!=-1&&!visit[down]) { p.pos=down; p.t=temp.t+1; q.push(p); } } return -1; } int main() { while(scanf("%d",&n),n) { scanf("%d %d",&start,&end); memset(floor,-1,sizeof(floor)); while(!q.empty()) q.pop(); for(int i=1;i<=n;i++) { int t; scanf("%d",&t); temp.pos=i; floor[i][0]=floor[i][1]=-1; if(i+t<=n) floor[i][0]=i+t; if(i-t>=1) floor[i][1]=i-t; if(i==start) { temp.t=0; q.push(temp); } } temp=q.front(); printf("%d\n",BFS()); } return 0; }
思路二：
DP: dp[i]表示从起点到第i层的按按钮的最少次数 31MS
DP：#include<cstdio> #include<cstring> #define maxn 250 #define INF 0x3f3f3f3f int dp[maxn],floor[maxn][2];//floor[i][0]表示第i层向上能到的楼层，floor[i][1]则表示向下能到的楼层 int main(){ int n,a,b; while(scanf("%d",&n),n){ scanf("%d %d",&a,&b); memset(floor,-1,sizeof(floor)); for(int i=1;i<=n;i++){ int t; scanf("%d",&t); if(i+t<=n) floor[i][0]=i+t; if(i-t>=1) floor[i][1]=i-t; } memset(dp,0x3f,sizeof(dp)); dp[a]=0; while(true){ int num=0;//记录更新数据的次数 for(int i=1;i<=n;i++){ if(dp[i]<INF){ for(int j=0;j<2;j++){ if(floor[i][j]!=-1){ if(dp[floor[i][j]]>dp[i]+1) dp[floor[i][j]]=dp[i]+1,num++; } } } } if(num==0)//无法继续更新 break; } if(dp[b]==INF) dp[b]=-1; printf("%d\n",dp[b]); } return 0; }