POJ 2481

 http://acm.pku.edu.cn/JudgeOnline/problem?id=2481
按S的升序、其次是E的降序进行排序，然后就跟POJ 2352 Stars做法一样了～～

  1 #include<stdio.h>
  2 #include<iostream>
  3 using namespace std;
  4 
  5 #include<math.h>
  6 #include<algorithm>
  7 #include<string.h>
  8 #include<stdlib.h>
  9 #include<vector>
 10 #include<set>
 11 #include<map>
 12 #include<stack>
 13 #include<string>
 14 #include<queue>
 15 
 16 #define repA(p,q,i)  for( int (i)=(p); (i)!=(q); ++(i) )
 17 #define repAE(p,q,i)  for( int (i)=(p); (i)<=(q); ++(i) )
 18 #define repD(p,q,i)  for( int (i)=(p); (i)!=(q); --(i) )
 19 #define repDE(p,q,i)  for( int (i)=(p); (i)>=(q); --(i) )
 20 #define range 1010
 21 #define end 100000
 22 
 23 struct node{
 24     int ID,x,y,num;
 25     node(){}
 26     node(int _ID, int _x, int _y)
 27         : ID(_ID), x(_x), y(_y) {}
 28 };
 29 
 30 int a[100010];
 31 node cow[100010];
 32 int n;
 33 
 34 bool cmp(node n1, node n2);
 35 bool cmp2(node n1, node n2);
 36 int lowbit(int i);
 37 void update(int loc);
 38 int sum(int loc);
 39 
 40 int main()
 41 {
 42     while(scanf("%d",&n) != EOF )
 43     {
 44         if(n==0)  break;
 45 
 46         int s,e;
 47         repA(0,n,i)
 48         {
 49             scanf("%d%d",&s,&e);
 50             cow[i] = node(i,e,s);
 51         }
 52 
 53         sort(cow, cow+n, cmp);
 54         //printf("\n");
 55         //repA(0,n,i)
 56           //printf("%d %d\n",cow[i].x, cow[i].y);
 57         memset(a, 0, sizeof(a) );
 58 
 59         node u;
 60         int tmp=0;
 61         update(cow[0].x);
 62         cow[0].num = 0;
 63         repA(1,n,i)
 64         {
 65             if(cow[i].x == cow[i-1].x && cow[i].y == cow[i-1].y)
 66               cow[i].num = cow[i-1].num;
 67             else cow[i].num = i - sum(cow[i].x-1) ;
 68             update(cow[i].x);
 69         }
 70 
 71         sort(cow, cow+n, cmp2);
 72 
 73         printf("%d",cow[0].num);
 74         repA(1,n,i)
 75             printf(" %d",cow[i].num);
 76         printf("\n");
 77 
 78     }
 79 }
 80 
 81 bool cmp(node n1, node n2)
 82 {
 83     if(n1.y != n2.y)
 84         return n1.y < n2.y ;
 85     else return n1.x > n2.x ;
 86 }
 87 
 88 bool cmp2(node n1, node n2)
 89 {
 90     return n1.ID < n2.ID ;
 91 }
 92 
 93 int lowbit(int i)
 94 {
 95     return i&(-i);
 96 }
 97 
 98 void update(int loc)
 99 {
100     if(loc == 0)  
101     {  ++a[0];  return;  }
102 
103     for(int i=loc; i<=end; i+=lowbit(i) )
104         a[i] += 1 ;
105     return ;
106 }
107 
108 int sum(int loc)
109 {
110     int all=a[0];
111     for(int i=loc; i>0; i-=lowbit(i) )
112         all += a[i];
113     return all;
114 }
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posted on 2013-11-20 16:51  码农之上~  阅读(259)  评论(0)  编辑  收藏  举报

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