求序列中满足Ai < Aj > Ak and i < j < k的组数 树状数组 HIT 2275 Number sequence

http://acm.hit.edu.cn/hoj/problem/view?id=2275

Number sequence

	Source : SCU Programming Contest 2006 Final
	Time limit : 1 sec		Memory limit : 64 M

Submitted : 1632, Accepted : 440

Given a number sequence which has N element(s), please calculate the number of different collocation for three number Ai, Aj, Ak, which satisfy that Ai < Aj > Ak and i < j < k.
Input

The first line is an integer N (N <= 50000). The second line contains N integer(s): A1, A2, ..., An(0 <= Ai <= 32768).
Output

There is only one number, which is the the number of different collocation.
Sample Input

5
1 2 3 4 1

Sample Output

6

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Statistic
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解法一

1.正序依次按如下步骤处理每个数据：用树状数组统计之前比它小的有多少，并记录在tmp数组里；从此数据开始向后更新更大的数。对此我要解释一下：树状数组的下标是数的大小，由于是按顺序来的，所以保证了i<j<k。由此可知道前面比此数小的有多少个。

2.将c数组清空后倒叙处理一遍数据。由此可知道后面比此数小的有多少个。

3.将每个数的两个数据相乘求和，最后用long long存储输出。

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#define MAXN 50010
#define MAXM 32771 
int a[MAXN];
int c[MAXM];
int tmp[MAXN];
int n;
int lowbit(int x){
    return x&(-x);
}
void add(int x,int ad){
    while(x<MAXM){
	    c[x]+=ad;
        x+=lowbit(x);
    }
}
int sum(int x){
    int res=0;
    while(x>0){
        res+=c[x];
        x-=lowbit(x);
    } 
    return res;
}
int main(){
    while(scanf("%d",&n)!=EOF){ 
	    long long ans=0;
	    memset(tmp,0,sizeof(tmp));
	    memset(c,0,sizeof(c));
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
      	    a[i]++;
           	tmp[i]=sum(a[i]-1);   //注意是严格的大于
            add(a[i],1); 
        }
        memset(c,0,sizeof(c));
        for(int i=n;i>=1;i--){
        	ans+=(long long)sum(a[i]-1)*tmp[i]; //(long long)很按理说重要，不过没有加也过了。 
        	add(a[i],1);
        }
        printf("%lld\n",ans);
    }
    return 0;
}

解法二

1 题目要求的是总共的搭配方式，满足Ai < Aj > Ak.并且i j k不同

2 我们开两个树状数组，第一个在输入的时候就去更新。然后我们在去枚举Aj 同时维护第二个树状数组，对于AI来说就是在第二个树状数组里面求和

然后在通过第一个树状数组就可以求出Ak的个数，把结果相乘即可

代码:

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

const int MAXN = 50010;

int n , num[MAXN];
int treeNumOne[MAXN];
int treeNumTwo[MAXN];

int lowbit(int x){
    return x&(-x);
}

int getSum(int *arr , int x){
    int sum = 0;
    while(x){
        sum += arr[x];
        x -= lowbit(x);
    }
    return sum;
}

void add(int *arr , int x , int val){
    while(x < MAXN){
         arr[x]  += val; 
         x += lowbit(x);
    }
}

long long getAns(){
    if(n < 3)
        return 0;
    long long ans = 0;
    add(treeNumTwo , num[1] , 1);
    for(int i = 2 ; i < n ; i++){
        int x = getSum(treeNumTwo , num[i]-1); 
        int y = getSum(treeNumOne , num[i]-1); 
        add(treeNumTwo , num[i] , 1);
        ans += (x)*(y-x); 
    }
    return ans;
}

int main(){
    while(scanf("%d" , &n) != EOF){
         memset(treeNumOne , 0 , sizeof(treeNumOne));
         memset(treeNumTwo , 0 , sizeof(treeNumTwo)); 
         for(int i = 1 ; i <= n ; i++){
             scanf("%d" , &num[i]);  
             num[i]++;
             add(treeNumOne , num[i] , 1);
         }
         printf("%lld\n" , getAns());
    }
    return 0;
}