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 }
