uva 10881 - Piotr's Ants （思维，3级）

Problem D
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 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 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 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, print "Turning" instead of "L" or "R" for their direction. If an ant falls off the pole before Tseconds, print "Fell off" for that ant. Print an empty line after each test case.

Sample InputSample Output

210 1 41 R5 R3 L10 R10 2 34 R5 L8 R

Case #1:2 Turning6 R2 TurningFell offCase #2:3 L6 R10 R

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Problemsetter: Igor Naverniouk
Alternate solutions: Frank Pok Man Chu and Yury Kholondyrev

思路：蚂蚁相遇做的操作实际上是交换两者的工作，所以只需确定最后所有蚂蚁的状态，以及对应的蚂蚁就行了。

  注意：注意最后输出是按照原始蚂蚁位置的排序。

#include<iostream>#include<cstring>#include<cstdio>#include<algorithm>using namespace std;const int mm=1e4+9;const char*out[]={"Fell off\n","Turning\n","L\n","R\n"};int cas;class node{  public:int pi,d,id;}f[mm],r[mm];bool cmp(node a,node b){  return a.pi<b.pi;}bool cmpp(node a,node b){  return a.id<b.id;}int zf[mm];int main(){ int L,T,n;char s;  while(~scanf("%d",&cas))  {    for(int ca=1;ca<=cas;++ca)    {      scanf("%d%d%d",&L,&T,&n);      for(int i=1;i<=n;++i)      {        scanf("%d %c",&f[i].pi,&s);        f[i].id=i;        if(s=='L')f[i].d=-1;        else f[i].d=1;        r[i].pi=f[i].pi;        r[i].id=i;        f[i].pi+=f[i].d*T;      }      f[0].pi=-1e9;      sort(r+1,r+n+1,cmp);      ///原始顺序对应的序列      for(int i=1;i<=n;++i)      zf[r[i].id]=i;      sort(f+1,f+n+1,cmp);      for(int i=1;i<=n;++i)      {        if(f[i].pi<0||f[i].pi>L)f[i].d=0;        else if(f[i].pi==f[i-1].pi)f[i-1].d=f[i].d=1;        else if(f[i].d==-1)f[i].d=2;        else if(f[i].d==1)f[i].d=3;        ///zf[r[i].id]=f[i].id;      }      int z;      ///sort(f+1,f+n+1,cmpp);      printf("Case #%d:\n",ca);      for(int i=1;i<=n;++i)      { z=zf[i];        if(f[z].d)        printf("%d ",f[z].pi);        printf("%s",out[f[z].d]);      }      puts("");    }  }}