UVA - 10881 Piotr's Ants

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 beforeT 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

水题，相撞后同时掉头就向对穿而过一样。。。只要维持一下相对顺序就行了

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>

using namespace std;

struct ANTS
{
    int st,ed,dir,num1;
    char output[120];
}ants[11000];

int L,T,N,order[11000];

bool cmp1(ANTS a,ANTS b)
{
    return a.st<b.st;
}

bool cmp2(ANTS a,ANTS b)
{
    return a.ed<b.ed;
}

int main()
{
    int t,cas=1;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d",&L,&T,&N);
        int x; char str[15];
        for(int i=0;i<N;i++)
        {
            scanf("%d%s",&x,str);
            ants[i].st=x;
            if(str[0]=='R') ants[i].dir=1;
            else ants[i].dir=-1;
            ants[i].ed=ants[i].dir*T+ants[i].st;
            ants[i].num1=i;
        }
        sort(ants,ants+N,cmp1);
        for(int i=0;i<N;i++)
        {
            order[ants[i].num1]=i;
        }
        sort(ants,ants+N,cmp2);
        printf("Case #%d:\n",cas++);
        for(int i=0;i<N;i++)
        {
            if(ants[i].ed>L||ants[i].ed<0)
            {
                sprintf(ants[i].output,"Fell off\n");
                continue;
            }
            if(i+1<N)
            {
                int next=i+1;
                if(ants[i].ed==ants[next].ed)
                {
                    sprintf(ants[i].output,"%d Turning\n",ants[i].ed);
                    i++;
                    sprintf(ants[i].output,"%d Turning\n",ants[next].ed);
                    continue;
                }
            }
            if(ants[i].dir==1)
            {
                sprintf(ants[i].output,"%d R\n",ants[i].ed);
            }
            else if(ants[i].dir==-1)
            {
                sprintf(ants[i].output,"%d L\n",ants[i].ed);
            }
        }
        for(int i=0;i<N;i++)
        {
            int ii=order[i];
            printf("%s",ants[ii].output);
        }
        putchar(10);
    }
    return 0;
}