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。。

 1 //STATUS:C++_AC_204MS_3048KB

 2 #include <functional>

 3 #include <algorithm>

 4 #include <iostream>

 5 //#include <ext/rope>

 6 #include <fstream>

 7 #include <sstream>

 8 #include <iomanip>

 9 #include <numeric>

10 #include <cstring>

11 #include <cassert>

12 #include <cstdio>

13 #include <string>

14 #include <vector>

15 #include <bitset>

16 #include <queue>

17 #include <stack>

18 #include <cmath>

19 #include <ctime>

20 #include <list>

21 #include <set>

22 //#include <map>

23 using namespace std;

24 //#pragma comment(linker,"/STACK:102400000,102400000")

25 //using namespace __gnu_cxx;

26 //define

27 #define pii pair<int,int>

28 #define mem(a,b) memset(a,b,sizeof(a))

29 #define lson l,mid,rt<<1

30 #define rson mid+1,r,rt<<1|1

31 #define PI acos(-1.0)

32 //typedef

33 typedef long long LL;

34 typedef unsigned long long ULL;

35 //const

36 const int N=100010;

37 const int INF=0x3f3f3f3f;

38 const int MOD=1e9+7,STA=8000010;

39 //const LL LNF=1LL<<60;

40 const double EPS=1e-8;

41 const double OO=1e15;

42 const int dx[4]={-1,0,1,0};

43 const int dy[4]={0,1,0,-1};

44 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};

45 //Daily Use ...

46 inline int sign(double x){return (x>EPS)-(x<-EPS);}

47 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}

48 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}

49 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}

50 template<class T> inline T Min(T a,T b){return a<b?a:b;}

51 template<class T> inline T Max(T a,T b){return a>b?a:b;}

52 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}

53 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}

54 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}

55 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}

56 //End

57 

58 struct Node{

59     int p,d;

60     bool operator < (const Node& a)const{

61         return p<a.p;

62     }

63 }be[N],af[N];

64 int L,n;

65 

66 int main()

67 {

68  //   freopen("in.txt","r",stdin);

69     int i,j,w,T;

70     char c;

71     while(~scanf("%d%d",&L,&n))

72     {

73         T=-INF;

74         for(i=0;i<n;i++){

75             scanf("%d %c",&w,&c);

76             be[i]=Node{w,c=='L'?-1:1};

77             if(c=='L')T=Max(T,w);

78             else T=Max(T,L-w);

79         }

80         for(i=0;i<n;i++)

81             af[i]=Node{be[i].p+be[i].d*T,0};

82 

83         sort(be,be+n);

84         sort(af,af+n);

85         for(i=0;i<n && af[i].p<0;i++);

86         if(af[i+1].p!=L)

87             printf("The last ant will fall down in %d seconds - started at %d.\n",T,be[i].p);

88         else printf("The last ant will fall down in %d seconds - started at %d and %d.\n",T,be[i].p,be[i+1].p);

89     }

90     return 0;

91 }