poj 3928 Ping pong

Ping pong

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4260		Accepted: 1407

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

Beijing 2008

首先简要说明题意，给定一个不存在重复数的序列，请问其中，有多少种可能，挑三个数出来，按位置排列，满足单增或者单减

首先我们先考虑，要求出每个数与它前面的数构成的顺序对数a，逆序对数b；与它后面的数构成的顺序对数c，逆序对数d

ans=a*c+b*d;

因为前面的一个顺序对与后面的一个顺序对 重合一个数，三个数，刚好满足题目的条件。

这里再解释一下如何求a,b,c,d：

先把所有数排序并且记录位置。

然后我们只需要将数从大到小，每个数所在的位置添加1，然后对于这个数的b和c只需要分别统计前缀和与后缀和即可

反向从小到大可以求出a，d

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=20005;
struct A
{
    int z,id;
    bool operator<(const A a)const
    {
        return z<a.z;
    }
}a[N];
int c[N],jl[N][2],n;
void add(int x)
{
    for(;x<=n;x+=x&-x)
        ++c[x];
}
int askk(int x)
{
    int re=0;
    for(;x;x-=x&-x)
        re+=c[x];
    return re;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        long long ans=0;
        scanf("%d",&n);
        for(int i=1;i<=n;++i)
            scanf("%d",&a[a[i].id=i].z);
        sort(a+1,a+n+1);
        memset(c,0,sizeof(c));
        for(int i=1;i<=n;++i)
        {
            add(a[i].id);
            jl[i][0]=askk(a[i].id-1);
            jl[i][1]=askk(n)-askk(a[i].id);
        }
        memset(c,0,sizeof(c));
        for(int i=n;i;--i)
        {
            add(a[i].id);
            ans+=(long long)jl[i][1]*askk(a[i].id-1);
            ans+=(long long)jl[i][0]*(askk(n)-askk(a[i].id));
        }
        printf("%lld
",ans);
    }
    return 0;
}