A strange lift
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8550 Accepted Submission(s): 3241
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
Recommend
8600
题意：n层楼。坐电梯。每层都有一个数字Ti，代表这层可以上或下Ti层。然后问能否从a层到b层。不能输出-1，能就输出最少需要几步。
感想：
1、我什么都不想说。。。诶。。。
虽然这个题数据量是挺大的。。。用BFS无悬念。但是我觉得用DFS写怎么着也应该是超时啊，怎么会是WA。。。？
2、用DFS写。由于电梯上下是并列的选择，
而dfs有搜索顺序的限制，不能找到就输出，所以不能保证第一次搜到的是最小的。只得用res1、res2记录下tot1、tot2的值。改变搜索顺序搜两次，
然后返回主函数比较最小值输出。两次dfs()有标记变量，记录找到没。如果都没找到，两个标记变量flag1、flag2就应该都是0，此时输出-1。
主函数中判断如果a==b就直接输出0，反之两次dfs()。
我写了以后怎么想都只会超时，为什么会wa呢。。。。
3、其实用bfs()写省去很多判断。比如没有顺序的限制，因为是层次遍历，而且第一个找到的一定是最小步数。比如不需要判断a==b，判断也可以，算是优化剪枝吧。。
bfsAC代码：
#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
using namespace std;
int t[210],vis[210];
int n,a,b;
struct node
{
int cur;
int steps;
}start;
int bfs()
{
queue<node> q;
while(!q.empty())
q.pop();
start.cur=a;
start.steps=0;
vis[a]=1;
q.push(start);
node m,temp;
while(!q.empty())
{
temp=q.front();
q.pop();
if(temp.cur==b)
return temp.steps;
int up=temp.cur+t[temp.cur];
if(!vis[up]&&up<=n)
{
vis[up]=1;
m.cur=up;
m.steps=temp.steps+1;
q.push(m);
}
int down=temp.cur-t[temp.cur];
if(!vis[down]&&down>0)
{
vis[down]=1;
m.cur=down;
m.steps=temp.steps+1;
q.push(m);
}
}
return -1;
}
int main()
{
while(scanf("%d",&n)&&n)
{
memset(vis,0,sizeof(vis));
scanf("%d%d",&a,&b);
for(int i=1;i<=n;i++)
scanf("%d",&t[i]);
printf("%d
",bfs());
}
return 0;
}
dfs写的wa代码，搞不懂怎么会wa。。。。超时我倒是想过。。。。
#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
int t[210];
int tot1,tot2,n,a,b;
bool vis[210];
int flag1,flag2;
int res1,res2;
void dfs1(int x)
{
//printf("now=%d
",x);
//printf("tot=%d
",tot);
if(flag1) return;
vis[x]=true;
if(x==b)
{
res1=tot1;//printf("%d
",tot);
flag1=1;
return;
}
else if(x>n||x<1) return;
else
{
if(x+t[x]<=n&&vis[x+t[x]]==false)
{
tot1++;
dfs1(x+t[x]);
tot1--;
if(flag1) return;
}
if(x-t[x]>=1&&vis[x-t[x]]==false)
{
tot1++;
dfs1(x-t[x]);
tot1--;
}
}
}
void dfs2(int x)
{
//printf("now=%d
",x);
if(flag2) return;
vis[x]=true;
if(x==b)
{
res2=tot2;
flag2=1;
return;
}
else if(x>n||x<1) return;
else
{
if(x-t[x]>=1&&vis[x-t[x]]==false)
{
tot2++;
dfs2(x-t[x]);
tot2--;
if(flag2) return;
}
if(x+t[x]<=n&&vis[x+t[x]]==false)
{
tot2++;
dfs2(x+t[x]);
tot2--;
}
}
}
int main()
{
int i;
while(scanf("%d",&n)&&n)
{
tot1=tot2=0,flag1=flag2=0;
res1=0,res2=0;
memset(vis,0,sizeof(vis));
scanf("%d%d",&a,&b);
for(i=1;i<=n;i++)
scanf("%d",&t[i]);
if(a==b) printf("0
");
else if(a>n||b>n) printf("-1
");
else
{
dfs1(a);
memset(vis,0,sizeof(vis));
dfs2(a);
if(flag1==0&&flag2==0) printf("-1
");
else if(flag1==0&&flag2) printf("%d
",res2);
else if(flag1&&flag2==0) printf("%d
",res1);
else printf("%d
",res2>=res1?res1:res2);
}
}
return 0;
}
/*
5 1 4
2 3 1 2 1
ans=2
5 1 5
2 3 1 2 1
ans=3
5 1 5
3 1 1 2 1
ans=-1
3 2 2
1 1 1
ans=0
*/