BNUOJ-26579 Bread Sorting YY

　　题目链接：http://www.bnuoj.com/bnuoj/problem_show.php?pid=26579

　　考虑两个性质：蚂蚁的相对位置不变，蚂蚁碰撞时相当于对穿而过，然后排两次序就可以了。。

 //STATUS:C++_AC_204MS_3048KB
 #include <functional>
 #include <algorithm>
 #include <iostream>
 //#include <ext/rope>
 #include <fstream>
 #include <sstream>
 #include <iomanip>
 #include <numeric>
 #include <cstring>
 #include <cassert>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <bitset>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <ctime>
 #include <list>
 #include <set>
 //#include <map>
 using namespace std;
 //#pragma comment(linker,"/STACK:102400000,102400000")
 //using namespace __gnu_cxx;
 //define
 #define pii pair<int,int>
 #define mem(a,b) memset(a,b,sizeof(a))
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define PI acos(-1.0)
 //typedef
 typedef long long LL;
 typedef unsigned long long ULL;
 //const
 const int N=;
 const int INF=0x3f3f3f3f;
 const int MOD=1e9+,STA=;
 //const LL LNF=1LL<<60;
 const double EPS=1e-;
 const double OO=1e15;
 const int dx[]={-,,,};
 const int dy[]={,,,-};
 const int day[]={,,,,,,,,,,,,};
 //Daily Use ...
 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 //End

 struct Node{
     int p,d;
     bool operator < (const Node& a)const{
         return p<a.p;
     }
 }be[N],af[N];
 int L,n;

 int main()
 {
  //   freopen("in.txt","r",stdin);
     int i,j,w,T;
     char c;
     while(~scanf("%d%d",&L,&n))
     {
         T=-INF;
         for(i=;i<n;i++){
             scanf("%d %c",&w,&c);
             be[i]=Node{w,c=='L'?-:};
             if(c=='L')T=Max(T,w);
             else T=Max(T,L-w);
         }
         for(i=;i<n;i++)
             af[i]=Node{be[i].p+be[i].d*T,};

         sort(be,be+n);
         sort(af,af+n);
         for(i=;i<n && af[i].p<;i++);
         if(af[i+].p!=L)
             printf("The last ant will fall down in %d seconds - started at %d.\n",T,be[i].p);
         else printf("The last ant will fall down in %d seconds - started at %d and %d.\n",T,be[i].p,be[i+].p);
     }
     return ;
 }