POJ 3928 Ping pong 树状数组

最新推荐文章于 2018-07-14 08:57:43 发布

MakingMaker

最新推荐文章于 2018-07-14 08:57:43 发布

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文章标签： integer distance insert each output input

本文链接：https://blog.youkuaiyun.com/MakingMaker/article/details/8039490

本文探讨了一种特殊的乒乓球比赛组织方式及其背后的算法实现。在一条东西走向的街道上居住着不同技能等级的选手，他们之间的比赛需遵循特定的规则：裁判的技能等级必须介于两名参赛者之间，且参赛者和裁判间的距离总和不得超过两者住所的距离。文章通过输入选手的技能等级分布来计算可能的比赛场次。

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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

Score:
#include<cstdio>
#include<string.h>
using namespace std;
typedef long long ll;
#define MAXN 100000
#define M 20006
int ah[MAXN],al[MAXN],lb[MAXN];
int a[M],r[M],l[M];

int sum(int b[MAXN],int k)
{
    int ans=0;
    while(k>0)
    {
        ans+=b[k];
        k-=lb[k];
    }
    return ans;
}

void insert(int b[MAXN],int k)
{
    while(k<=MAXN)
    {
        b[k]++;
        k+=lb[k];
    }
}
int main()
{
    int T;
    scanf("%d",&T);
    for(int i=0;i<=MAXN;i++)
    {
        lb[i]=i&(-i);
    }
    while(T--)
    {
        int n,i;
        scanf("%d",&n);
        for(i=0;i<n;i++)
        {
            scanf("%d",a+i);
        }
        memset(ah,0,sizeof(ah));
        memset(al,0,sizeof(al));
        for(i=0;i<n;i++)
        {
            l[i]=sum(al,a[i]-1);
            insert(al,a[i]);
        }
        for(i=n-1;i>=0;i--)
        {
            r[i]=sum(ah,a[i]-1);
            insert(ah,a[i]);
        }
        ll ans=0;
        for(i=0;i<n;i++)
        {
            ans+=(ll)l[i] * (ll)( n - i - 1 - r[i] ) + (ll)( i - l[i] ) * (ll)r[i];
        }
        printf("%lld\n",ans);
    }
    return 0;
}