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=100010;
const int INF=0x3f3f3f3f;
const int MOD=1e9+7,STA=8000010;
//const LL LNF=1LL<<60;
const double EPS=1e-8;
const double OO=1e15;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//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=0;i<n;i++){
            scanf("%d %c",&w,&c);
            be[i]=Node{w,c=='L'?-1:1};
            if(c=='L')T=Max(T,w);
            else T=Max(T,L-w);
        }
        for(i=0;i<n;i++)
            af[i]=Node{be[i].p+be[i].d*T,0};

        sort(be,be+n);
        sort(af,af+n);
        for(i=0;i<n && af[i].p<0;i++);
        if(af[i+1].p!=L)
            printf("The last ant will fall down in %d seconds - started at %d.
",T,be[i].p);
        else printf("The last ant will fall down in %d seconds - started at %d and %d.
",T,be[i].p,be[i+1].p);
    }
    return 0;
}