【树状数组--思维】poj 3928 pingpong

Ping pong

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 3476 Accepted: 1257

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

题意：要举行比赛的话，需要一个裁判，两个选手，裁判位置在两位选手之间，技能值也要在两位选手之间；先从左到右枚举每个位置i，统计左侧技能值比他低的人数L[i]，再从右到左枚举每个位置i，统计右侧技能值比他低的人数R[i];这样对于每个位置i就有L[i]*(n-i-R[i])+(i-L[i]-1)*R[i]常比赛；

代码：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;typedef long long ll;const int N=100005;ll A[N],C[N];ll L[N],R[N];int n;int lowbit(int x){    return x&(-x);}ll sum(int x){    ll ret=0;    while(x>0)    {        ret+=C[x];        x-=lowbit(x);    }    return ret;}void add(int x,int d){    while(x<=N)    {        C[x]+=d;        x+=lowbit(x);    }}int main(){    int T;    scanf("%d",&T);    while(T--)    {        memset(A,0,sizeof(A));        int n;        scanf("%d",&n);        memset(C,0,sizeof(C));        for(int i=1;i<=n;i++)        {            scanf("%lld",&A[i]);            add(A[i],1);            L[i]=sum(A[i]-1);        }        memset(C,0,sizeof(C));        for(int i=n;i>=1;i--)        {            add(A[i],1);            R[i]=sum(A[i]-1);        }//        for(int i=1;i<=n;i++)//        {//            printf("L[i]:%lld R[i]:%lld\n",L[i],R[i]);//        }        ll sum=0;        for(int i=1;i<=n;i++)            sum+=L[i]*(n-i-R[i])+(i-L[i]-1)*R[i];        printf("%lld\n",sum);    }    return 0;}