HDU 2492 Ping pong (樹狀數組)

題目鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=2492

Ping pong

Problem 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

話不多說直接上代碼

解釋都在代碼中標注了

 1 #include<stdio.h>
 2 #include<string.h>
 3 #define size 100100
 4 int n,c[size],x1[size],x2[size],y1[size],y2[size],a[size];
 5 
 6 int Lowbit(int k)
 7 {
 8     return (k&-k);
 9 }
10 void update(int pos,int num)
11 {
12     while(pos<size)//重要  是size  而不是<=n
13     {
14         c[pos]+=num;
15         pos+=Lowbit(pos);
16     }
17 }
18 int sum(int pos)
19 {
20     int s=0;
21     while(pos>0)
22     {
23         s+=c[pos];
24         pos-=Lowbit(pos);
25     }
26     return s;
27 }
28 
29 
30 
31 int main()
32 {
33        int i,cas;
34        scanf("%d",&cas);
35        while(cas--)
36        {
37            memset(c,0,sizeof(c));
38            scanf("%d",&n);
39            for(i=1;i<=n;i++) 
40            {
41                scanf("%d",&a[i]);   
42                int k1; 
43                k1=sum(a[i]);
44                x1[i]=k1; //輸入的i個數中 有k1個比a[i]小
45                x2[i]=i-1-k1; //輸入的i個數中 有k1個比a[i]大
46                update(a[i],1);
47            }
48            memset(c,0,sizeof(c));
49            int j=1;//代表現在輸入的數的個數 
50            for(i=n;i>=1;i--,j++) 
51            {
52                  int k1;
53                  k1=sum(a[i]);
54                  y1[i]=k1;//輸入a[i]后輸入的那些數中有多少個比a[i]小的
55                  y2[i]=j-1-k1; //輸入a[i]后輸入的那些數中有多少個比a[i]大的
56                  update(a[i],1);
57            }
58            __int64 ans=0;
59            for(i=1;i<=n;i++)
60            {
61               // printf("x1[%d]=%d x2[%d]=%d y1[%d]=%d  y2[%d]=%d\n",i,x1[i],i,x2[i],i,y1[i],i,y2[i]);
62               ans+=x1[i]*y2[i]+x2[i]*y1[i];
63                // ans+=x1[i]*x2[i];
64            }
65            printf("%I64d\n",ans);
66        }
67 
68 }

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posted @ 2014-08-15 19:41 四十四次日落 Views(406) Comments(0) 收藏舉報

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