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

51 2 3 4 1

Sample Output

6

• Submit
• Status
• Statistic
• Solution
• Discuss
• Print

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;}