HDOJ 5523 Game



Game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 387    Accepted Submission(s): 150


[align=left]Problem Description[/align]
XY is playing a game:there are N pillar in a row,which numbered from 1 to n.Each pillar has a jewel.Now XY is standing on the S-th pillar and the exit is in the T-th pillar.XY can leave from the exit only
after they get all the jewels.Each time XY can move to adjacent pillar,or he can jump to boundary ( the first pillar or the N-th pillar) by using his superpower.However,he needs to follow a rule:if he left the pillar,he no can not get here anymore.In order
to save his power,XY wants to use the minimum number of superpower to pass the game.
 

[align=left]Input[/align]
There are multiple test cases, no more than 1000 cases.

For each case,the line contains three integers:N,S and T.(1≤N≤10000,1≤S,T≤N)
 

[align=left]Output[/align]
The output of each case will be a single integer on a line: the minimum number of using superpower or output -1 if he can't leave.
 

[align=left]Sample Input[/align]

4 1 4
4 1 3

 

[align=left]Sample Output[/align]

0
1

 

题意： 有n根柱子排成一排，起始位置是第S根柱子，出口位置是第T根柱子。要求把所有的柱子都走一遍，且走过的不能再走。走法有两种：1，走到与当前相邻的柱子上;    2，直接跳到第一根或者最后一根柱子上，并消耗能量。 

问最少消耗多少能量把柱子走完，若不能把柱子走完就输出-1

题解：很容易就知道能把柱子走完的情况下，最多消耗2个能量。  当n不等于1时，S==T时就无法从T根柱子结束（走过的柱子不能再走），所以此种情况下是不能完成的。仅有此一种情况输出-1。   其他各种情况分别讨论即可。

代码如下：

#include<cstdio>
#include<cstring>
int main()
{
int n,s,t;
while(scanf("%d%d%d",&n,&s,&t)!=EOF)
{
if(n==1)
{
printf("0\n");
continue;
}
if(s==t&&n!=1)
{
printf("-1\n");
continue;
}
if((s==1&&t==n)||(s==n&&t==1))
{
printf("0\n");
continue;
}
if(s==1||s==n)
{
printf("1\n");
continue;
}
if(s==t+1||s==t-1)
{
printf("1\n");
continue;
}
printf("2\n");
}
return 0;
}

BestCoder上的题目算法考察不多，不过挺锻炼思维能力的。以后常玩