码迷,mamicode.com
首页 > 其他好文 > 详细

UVA-10881-Piotr's Ants

时间:2015-01-26 11:34:02      阅读:150      评论:0      收藏:0      [点我收藏+]

标签:

题目:UVA-10881-Piotr‘s Ants

思路:模拟,看成对穿而过求终状态,然而相对顺序不变,排序即可。

 1 #include <iostream>
 2 #include<stdio.h>
 3 #include<stdlib.h>
 4 #include<string.h>
 5 #include<algorithm>
 6 #include<math.h>
 7 
 8 using namespace std;
 9 
10 const char dirname[3][10]= {"L","Turning","R"};
11 int order[10010];
12 
13 struct ant
14 {
15     int id,position,direct;
16 } before[10010],after[10010];
17 
18 bool cmp(ant x,ant y)
19 {
20     return x.position<y.position;
21 }
22 
23 int main()
24 {
25     int c,cnt;
26     scanf("%d",&c);
27     for(cnt=1; cnt<=c; cnt++)
28     {
29         int t,l,n,i,tmp;
30         printf("Case #%d:\n",cnt);
31         scanf("%d%d%d",&l,&t,&n);
32         for(i=0; i<n; i++)
33         {
34             int p,d;
35             char dir;
36             scanf("%d %c",&p,&dir); //#尼玛中间要空格呀!!!
37             d=(dir==L?-1:1); //#这样求终态较方便
38             before[i]=(ant)
39             {
40                 i,p,d
41             };
42             after[i]=(ant)
43             {
44                 0,p+d*t,d
45             };
46         }
47         sort(before,before+n,cmp);
48         for(i=0; i<n; i++)
49             order[before[i].id]=i;
50         sort(after,after+n,cmp);
51         for(i=0; i<n-1; i++)
52             if(after[i].position==after[i+1].position)
53                 after[i].direct=after[i+1].direct=0;
54         for(i=0; i<n; i++)
55         {
56             tmp=order[i];
57             if(after[tmp].position<0||after[tmp].position>l) printf("Fell off\n");
58             else printf("%d %s\n",after[tmp].position,dirname[after[tmp].direct+1]);
59         }
60         printf("\n");
61     }
62     return 0;
63 }

 

UVA-10881-Piotr's Ants

标签:

原文地址:http://www.cnblogs.com/alohagin/p/4249713.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!