uva 10881 Piotr's Ants 模拟

碰撞后相当于穿过去，所以结束的位置是可以确定的，又因为相对位置不变，所以可以通过排序确定某个位置是第几只蚂蚁。

#include<iostream>
#include<cmath>
#include<cstdio>
#include<sstream>
#include<cstdlib>
#include<string>
#include<string.h>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<ctime>
#include<bitset>
#define eps 1e-6
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
//#define ll __int64
#define ll long long
#define lson l,m,(rt<<1)
#define rson m+1,r,(rt<<1)|1
#define M 1000000007
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
int l,t,n;
struct node
{
    int id;
    int st;
    string dir;
}ant[100005],now[100005];
int p[100005];
bool cmp1(node x,node y)
{
    return x.st<y.st;
}
int main()
{
    int cas,ca=1,n;
    scanf("%d",&cas);
    while(cas--)
    {
        scanf("%d%d%d",&l,&t,&n);
        for(int i=0;i<n;i++)
        {
            cin>>ant[i].st>>ant[i].dir;
            ant[i].id=i;

            now[i].st=ant[i].st+(ant[i].dir=="L"?-1*t:t);
            now[i].dir=ant[i].dir;
        }
        sort(ant,ant+n,cmp1);
        for(int i=0;i<n;i++)
        {
            p[ant[i].id]=i;
        }
        sort(now,now+n,cmp1);
        for(int i=0;i<n-1;i++)
        {
            if(now[i].st==now[i+1].st) now[i].dir=now[i+1].dir="Turning";
        }
        printf("Case #%d:\n",ca++);
        for(int i=0;i<n;i++)
        {
           int id=p[i];
           if(now[id].st<0||now[id].st>l) puts("Fell off");
           else cout<<now[id].st<<" "<<now[id].dir<<endl;
        }
        puts("");
    }
    return 0;
}