Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
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"?
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
3
题目大意：
有一个电梯，在不同的楼层，可以上升或下降的层数是不同的
（例如第一层有一个整数3，说明他可以上升3层或者下降3层），
现在给你一个起始位置和最终位置，问至少需要几步能够到达最终位置。
解题思路：
该题可以用最短路，也可以用广搜来写
最短路：需要注意的是这题是单向边
bfs：每次只需要搜索两个方向
最短路算法：
1 #include <stdio.h> 2 #include <string.h> 3 #include <algorithm> 4 using namespace std; 5 6 const int inf = 1<<30; 7 8 int n; 9 int map[205][205]; 10 int a[205],cnt; 11 int vis[205],cast[205]; 12 13 void Dijkstra(int s,int e) //迪杰斯特拉 14 { 15 int i,j,min,pos; 16 memset(vis,0,sizeof(vis)); 17 for(i = 0; i<n; i++) 18 cast[i] = map[s][i]; 19 cast[s] = 0; 20 vis[s] = 1; 21 for(i = 1; i<n; i++) 22 { 23 min = inf; 24 for(j = 0; j<n; j++) 25 { 26 if(cast[j]<min && !vis[j]) 27 { 28 pos = j; 29 min = cast[j]; 30 } 31 } 32 if(min == inf) 33 break; 34 vis[pos] = 1; 35 for(j = 0; j<n; j++) 36 { 37 if(cast[pos]+map[pos][j]<cast[j] && !vis[j]) 38 cast[j] = cast[pos]+map[pos][j]; 39 } 40 } 41 } 42 43 int main() 44 { 45 int i,j,s,e,x,y; 46 while(~scanf("%d",&n),n) 47 { 48 scanf("%d%d",&s,&e); 49 s--,e--; 50 for(i = 0; i<n; i++) 51 for(j = 0; j<n; j++) 52 map[i][j] = inf; 53 for(i = 0; i<n; i++) //单向边 54 { 55 scanf("%d",&a[i]); 56 if(i+a[i]<n) //上升 57 map[i][i+a[i]] = 1; 58 if(i-a[i]>=0) //下降 59 map[i][i-a[i]] = 1; 60 } 61 Dijkstra(s,e); 62 printf("%d ",cast[e]==inf?-1:cast[e]); 63 } 64 65 return 0; 66 }
广搜 bfs：
1 #include <stdio.h> 2 #include <string.h> 3 #include <algorithm> 4 #include <queue> 5 using namespace std; 6 7 int a[210]; // 在第i层可以上升和下降的层数 8 int f[210]; // 记录步数 9 int n,x,y; 10 void bfs() 11 { 12 int b; 13 memset(f,0,sizeof(f)); 14 queue<int > q; 15 q.push(x); 16 f[x] = 1; 17 if (x == y) 18 return ; 19 while (!q.empty()) 20 { 21 b = q.front(); 22 if (b+a[b] <= n && !f[b+a[b]]) // 判断上升之后是否满足条件 23 { 24 q.push(b+a[b]); 25 f[b+a[b]] = f[b] + 1; 26 if (b+a[b] == y) // 到达之后直接跳出 27 return ; 28 } 29 if (b-a[b] > 0 && !f[b-a[b]]) // 判断下降之后是否满足条件 30 { 31 q.push(b-a[b]); 32 f[b-a[b]] = f[b] + 1; 33 if (b-a[b] == y) // 到达之后直接跳出 34 return ; 35 } 36 q.pop(); 37 } 38 return ; 39 } 40 int main () 41 { 42 int i,j; 43 while (scanf("%d",&n),n) 44 { 45 scanf("%d%d",&x,&y); 46 for (i = 1; i <= n; i ++) 47 scanf("%d",&a[i]); 48 49 bfs(); 50 if (f[y] != 0) 51 printf("%d ",f[y]-1); 52 else 53 printf("-1 "); 54 } 55 return 0; 56 }