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  • 比赛之树状数组题

     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 #include <algorithm>
     5 using namespace std;
     6 #define F first
     7 #define S second
     8 #define MP(a,b) make_pair( make_pair(a,b) , -1 )
     9 typedef pair< pair<int,int> , int> coder ;
    10 const int maxn = 300010 ;
    11 const int N = 100000 ;
    12 
    13 int sum[N+1] , ans[maxn] ;
    14 coder p[maxn] ;
    15 
    16 inline int lowbit(int x){
    17     return x & -x ;
    18 }
    19 void update(int x) {
    20     for(; x<=N; x+=lowbit(x))  sum[x]++ ;
    21 }
    22 int query(int x) {
    23     int ret = 0 ;
    24     for(; x>0 ; x-=lowbit(x)) ret += sum[x] ;
    25     return ret ;
    26 }
    27 
    28 int main()
    29 {
    30     int n ;
    31     scanf("%d", &n);
    32     for(int i=1; i<=n; i++)
    33         scanf("%d%d", &p[i].F.F , &p[i].F.S) , p[i].S = i ;
    34     sort(p+1 , p+1+n) ;
    35     int y = 1;
    36 
    37     for(int i=1; i<=n; i++) {
    38         int u = p[i].S ;
    39         while(y<i && p[y].F.F < p[i].F.F) {
    40             update(p[y++].F.S) ;
    41         }
    42         ans[u] += query(p[i].F.S) ;
    43         if(y < i) {
    44             ans[u] += lower_bound(p+y , p+i , MP(p[i].F.F , p[i].F.S)) - p  - y ;
    45         }
    46     }
    47 
    48     for(int i=1; i<=n; i++)
    49         printf("%d
    " , ans[i]) ;
    50 
    51     return 0;
    52 }
    AC参考代码
    题目:
    C - Coder Ratings
    Time Limit:1000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

    Description

    Some of the more elite (and not-so-elite) coders around take part in a certain unnamed programming contest. In said contest, there are multiple types of competitions. Here, we consider the Open and High School competition types. For each type, each competitor receives a rating, an integer between 1 and 100000, inclusive. A coder's rating is based upon his or her level of performance in matches and is calculated using a complicated formula which, thankfully, you will not be asked to implement.

    Although the Open and High School ratings for a coder who has participated in both competition types lately are usually close, this is not always the case. In particular, High School matches are more about speed, since many coders are able to solve all the problems, whereas Open matches require more thinking and there is a steeper curve in terms of problem difficulty.

    Problem Statement
    You are given N coders (1 ≤ N ≤ 300000), conveniently numbered from 1 to N. Each of these coders participates in both High School and Open matches. For each coder, you are also given an Open rating Ai and a High School rating Hi. Coder i is said to be better than coder j if and only if both of coder i's ratings are greater than or equal to coder j's corresponding ratings, with at least one being greater. For each coder i, determine how many coders coder i is better than.

    Input Format
    On the first line of input is a single integer N, as described above.
    N lines then follow. Line i+1 contains two space-separated integers, Ai and Hi.

    Output Format
    Line i should contain the number of coders that coder i is better than.

    Sample Input

    8
    1798 1832
    862 700
    1075 1089
    1568 1557
    2575 1984
    1033 950
    1656 1649
    1014 1473
    

    Sample Output

    6
    0
    2
    4
    7
    1
    5
    1
    我要坚持一年,一年后的成功才是我想要的。
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  • 原文地址:https://www.cnblogs.com/tianxia2s/p/3901315.html
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