uva1428 Ping pong

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 rst 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 rst integer is N , the number of players. Then N distinct integers a 1 ; a 2 : : : a N follow, indicating the skill
rank of each player, in the order of west to east (1  a i  100000, i
= 1 : : : N ). Output For each test case, output a single line contains an integer, the total number of different games.

考虑每个人当裁判，需要知道他前面和后面分别有多少人比他大和比他小。扫描两遍用树状数组维护。

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define LL long longconst int maxm=100000;int s[100010],a[20010],n,f[20010],g[20010];int qry(int x){    int ret=0;    for (;x;x-=x&-x)      ret+=s[x];    return ret;}void add(int x){    for (;x<=maxm;x+=x&-x)      s[x]++;}int main(){    int i,T;    LL ans;    scanf("%d",&T);    while (T--)    {        scanf("%d",&n);        for (i=1;i<=n;i++)          scanf("%d",&a[i]);        memset(s,0,sizeof(s));        for (i=1;i<=n;i++)        {            f[i]=qry(a[i]-1);            add(a[i]);        }        memset(s,0,sizeof(s));        for (i=n;i;i--)        {            g[i]=qry(a[i]-1);            add(a[i]);        }        ans=0;        for (i=1;i<=n;i++)          ans+=(LL)f[i]*(n-i-g[i])+(LL)g[i]*(i-f[i]-1);        printf("%lld\n",ans);    }}