UVA 10881 Piortr‘s Ants 思维 模拟

Piotr's Ants
Time Limit: 2 seconds

"One thing is for certain: there is no stopping them;
the ants will soon be here. And I, for one, welcome our
new insect overlords."

Kent Brockman

Piotr likes playing with ants. He has n of them on a horizontal poleL cm long. Each ant is facing either left or right and walks at a constant speed of 1 cm/s. When two ants bump into each other, they both turn around (instantaneously) and start walking in opposite directions. Piotr knows where each of the ants starts and which direction it is facing and wants to calculate where the ants will end upTseconds from now.

Input
The first line of input gives the number of cases, N. N test cases follow. Each one starts with a line containing 3 integers:L ,T and n(0 <= n <= 10000). The nextn lines give the locations of then ants (measured in cm from the left end of the pole) and the direction they are facing (L or R).

Output
For each test case, output one line containing "Case #x:" followed byn lines describing the locations and directions of then ants in the same format and order as in the input. If two or more ants are at the same location, print "Turning" instead of "L" or "R" for their direction. If an ant falls off the polebefore T seconds, print "Fell off" for that ant. Print an empty line after each test case.

Sample Input	Sample Output
2 10 1 4 1 R 5 R 3 L 10 R 10 2 3 4 R 5 L 8 R	Case #1: 2 Turning 6 R 2 Turning Fell off Case #2: 3 L 6 R 10 R

---

Problemsetter: Igor Naverniouk

Alternate solutions: Frank Pok Man Chu and Yury Kholondyrev

可以直接看作蚂蚁的单向运动，emmmm

感觉问题主要是输出的顺序，蚂蚁的相对顺序是不变的，在pos改变后依旧需要按原来的相对顺序输出，这里用order记录

/*↑↑这个用法之前没用过呢，记笔记记笔记*/

RE了好多次……结果发现order开小了orz，还有别忘了输出空行……

 

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<cmath>
#include<algorithm>
typedef long long ll;
using namespace std;
struct ants{
    int pos;
    char dir;
    int id;
    int tu=0;
};
int n,m,l,t,T;
ants b[10005];
int order[10005];
bool cmp2(ants a,ants b)
{
    return  a.pos<b.pos;
} 
int main()
{
    scanf("%d",&t);
    for(int j=1;j<=t;j++)
    {
        scanf("%d%d%d",&l,&T,&m);
        for(int i=0;i<m;i++)
        {
          cin>>b[i].pos>>b[i].dir;
          b[i].id=i;
        }
        printf("Case #%d:
",j);
        sort(b,b+m,cmp2);
        for(int i=0;i<m;i++) //相对顺序处理 
        {
        order[b[i].id]=i;
         b[i].tu=0;
         if(b[i].dir=='R')
          b[i].pos=b[i].pos+T;
          else b[i].pos=b[i].pos-T;
        }
        sort(b,b+m,cmp2);
        for(int i=0;i<m-1;i++)
        {
            if(b[i].pos==b[i+1].pos&&b[i].dir!=b[i+1].dir)
             {
                 b[i].tu=1;
                 b[i+1].tu=1;
             }
        }
        for(int i=0;i<m;i++)
        {
            int ord=order[i];
            if(b[ord].tu)
            printf("%d Turning
",b[ord].pos);
            else if(b[ord].pos<0||b[ord].pos>l)
            printf("Fell off
");
            else 
            printf("%d %c
",b[ord].pos,b[ord].dir);
        }
        puts("");
     } 
    return 0;
}