[UVA10881][模拟]Piotr's Ants

\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 pole L 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 up T
seconds from now.
Input
The rst 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 next n lines give the locations of the n 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 by n lines describing the locations
and directions of the n ants in the same format and order as in the input. If two or more ants are at
the same location, printTurning’ instead of L' orR’ for their direction. If an ant falls off the pole
before T seconds, print `Fell off’ for that ant. Print an empty line after each test case.
Sample Input
2
10 1 4
1 R
5 R
3 L
10 R
10 2 3
4 R
5 L
8 R
Sample Output
Case #1:
2 Turning
6 R
2 Turning
Fell off
Case #2:
3 L
6 R
10 R

国庆期间考过一道由这题改编过来的题，这道题没有什么算法可言，主要是考察对题目的分析
首先由于蚂蚁相撞就会反向，所有每一个蚂蚁的相对位置关系是恒定不变的。其次，对于连个相撞的蚂蚁，我们可以看做是它们互相穿过，所以如果一个蚂蚁在不相撞的情况下能够走到pos，那么最终一定会有一只蚂蚁走到pos。
所以思路就来了，我们需要计算出每一个蚂蚁理想情况的末位置，然后按照蚂蚁开始的相对位置，将蚂蚁与末位置一一匹配即可

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn=1e4+5;int K,L,T,n,order[maxn];const char DIR[4][10]={"L","Turning","R"};struct ant{    int ID,pos,dir;    bool operator<(const ant &x)const     {        return pos<x.pos;    }}before[maxn],after[maxn];int main(){//  freopen("ant.in","r",stdin);    scanf("%d",&K);    int k=K;    while (k--)    {        printf("Case #%d:\n",K-k);        scanf("%d%d%d",&L,&T,&n);        for (int i=0;i<n;i++)        {            int pos,dir;            char ch;            scanf("%d %c",&pos,&ch);            dir=ch=='R'?1:-1;            before[i]=(ant){i,pos,dir};            after[i]=(ant){0,pos+T*dir,dir};        }        sort(before,before+n);        for (int i=0;i<n;i++) order[before[i].ID]=i;        sort(after,after+n);        for (int i=0;i<n-1;i++)          if (after[i].pos==after[i+1].pos) after[i].dir=after[i+1].dir=0;        for (int i=0;i<n;i++)        {            int temp=order[i];            if (after[temp].pos<0||after[temp].pos>L) printf("Fell off\n");            else printf("%d %s\n",after[temp].pos,DIR[after[temp].dir+1]);        }        printf("\n");    }    return 0;}