zoukankan      html  css  js  c++  java
  • POJ3928Ping pong[树状数组 仿逆序对]

    Ping pong
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 3109   Accepted: 1148

    Description

    N(3<=N<=20000) ping pong players live along a west-east street(consider the street as a line segment). Each player has a unique skill rank. To improve their skill rank, they often compete with each other. If two players want to compete, they must choose a referee among other ping pong players and hold the game in the referee's house. For some reason, the contestants can't choose a referee whose skill rank is higher or lower than both of theirs. The contestants have to walk to the referee's house, and because they are lazy, they want to make their total walking distance no more than the distance between their houses. Of course all players live in different houses and the position of their houses are all different. If the referee or any of the two contestants is different, we call two games different. Now is the problem: how many different games can be held in this ping pong street?

    Input

    The first line of the input contains an integer T(1<=T<=20), indicating the number of test cases, followed by T lines each of which describes a test case. 
    Every test case consists of N + 1 integers. The first integer is N, the number of players. Then N distinct integers a1, a2 ... aN follow, indicating the skill rank of each player, in the order of west to east. (1 <= ai <= 100000, i = 1 ... N).

    Output

    For each test case, output a single line contains an integer, the total number of different games. 

    Sample Input

    1 
    3 1 2 3

    Sample Output

    1

    Source


    白书
    求f[i]为i之前技能值小于i的人数,g[i]为i之后小于i的人数
    求法类似逆序对,x[i]技能值为i的人是否有,用树状数组维护x,求和即可
    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    using namespace std;
    typedef long long ll;
    const int N=2e4+5,M=1e5+5;
    inline int read(){
        char c=getchar();int x=0,f=1;
        while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
        while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
        return x*f;
    }
    int T,n,m,a[N],bit[M],f[N],g[N];
    inline int lowbit(int x){return x&-x;}
    inline void add(int p,int v,int bit[]){
        for(int i=p;i<=m;i+=lowbit(i)) bit[i]+=v; 
    }
    inline int sum(int p,int bit[]){
        int ans=0;
        for(int i=p;i>0;i-=lowbit(i)) ans+=bit[i];
        return ans;
    }
    int main(){
        T=read();
        while(T--){
            n=read();
            for(int i=1;i<=n;i++) a[i]=read(),m=max(m,a[i]);
            memset(bit,0,sizeof(bit));
            for(int i=1;i<=n;i++){
                add(a[i],1,bit);
                f[i]=sum(a[i]-1,bit);
            }
            memset(bit,0,sizeof(bit));
            for(int i=n;i>=1;i--){
                add(a[i],1,bit);
                g[i]=sum(a[i]-1,bit);
            }
            ll ans=0;
            for(int i=2;i<=n-1;i++) ans+=f[i]*(n-i-g[i])+g[i]*(i-1-f[i]);//,printf("t%d %d %d
    ",i,f[i],g[i]);
            printf("%lld
    ",ans);
        }
    }
  • 相关阅读:
    solrCloud设置Tomcat jvm内存解决内存溢出的问题
    JAVA 将图片转换为Base64编码
    希望这是一个新的开始,也是一个好的开端
    最全华为鸿蒙 HarmonyOS 开发资料汇总
    iPhone X适配方案
    前端程序员经常忽视的一个JavaScript面试题
    vs for Mac 升级后编译原项目提示找不到“.NETFramework,Version=v5.0”的引用程序集
    使用FastReport的BarCode2D控件生成含中文的PDF417条形码
    vs for Mac中的启用Entity Framework Core .NET命令行工具
    64位Windows7升级IE11后无法启动的解决办法
  • 原文地址:https://www.cnblogs.com/candy99/p/5962425.html
Copyright © 2011-2022 走看看