树状数组 LA 4329 亚洲赛北京赛区题

Posted 2020-09-27 lxjshuju

复习下树状数组

还是蛮有意思的一道题：

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=501&page=show_problem&problem=4174

学到几点：
1、树状数组C[i]的构建，一则c[i]=s[i]-s[i-lowbit(i)];这是一直用的做法。如今学到一种新的，直接add(i,a[i])，（s[i]为a[1]到a[i]的和）

2、前缀和思想，树状数组的Sum本身就是基于前缀和的思想。本题把比某数小的数的个数，通过开大量空间+后缀数组，高效的统计出来比某数小的数的个数

3、事实上我认为通过这个题。能够做出来一种O(nlogn)的排序算法。当然不完好的地方就是仅仅能是整数了，可是应该能够用vector+map解决？

贴自己的代码先。，，由于后面的Lrj大牛的代码太简洁...

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <functional>

using namespace std;

#define MAXN 20010
#define SIZE 100015

int a[MAXN],sma[SIZE],c[SIZE],s[SIZE],e[SIZE],tot[SIZE];
int n,mmax;

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

int sum(int i)
{
    int ans=0;
    for(;i>0;i-=lowbit(i))
        ans+=c[i];
    return ans;
}

  void add(int x, int d) {
    while(x <= SIZE) {
      c[x] += d; x += lowbit(x);
    }
  }

void Init()
{
        memset(c,0,sizeof(c));
        memset(a,0,sizeof(a));
        memset(sma,0,sizeof(sma));
        memset(s,0,sizeof(s));
        memset(e,0,sizeof(e));
        memset(tot,0,sizeof(tot));
}

int main()
{
   //freopen("la4329.txt","r",stdin);
    int t;
    long long ans;

    scanf("%d",&t);

    while(t--)
    {
        mmax=0;
        ans=0;
        Init();

        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            mmax=max(mmax,a[i]);
            add(a[i],1);
            //c[i]=sum(a[i]-1);
            e[a[i]]=1;
            sma[a[i]]=sum(a[i]-1);

        }
        memset(c,0,sizeof(c));
        for(int i=1;i<=mmax;i++)
        {
            s[i] =s[i-1]+e[i];
            c[i]=s[i]-s[i-lowbit(i)];
            tot[i]=sum(i-1);
        }
        ///////////////////////
        //for(int i=1;i<=n;i++)
        //{
        //    printf("sma[a[%d]] = %d  tot(%d) = %d\n",i,sma[a[i]],a[i],tot[a[i]]);
        //}
        ///////////////////////
        for(int i=1;i<=n;i++)
        {
            int tmp = tot[a[i]]-sma[a[i]];/*a[i]之后比a[i]小的个数*/
            ans+=(long long )tmp*(i-1-sma[a[i]])+(long long)sma[a[i]]*(n-i-tmp);
        }

        printf("%lld\n",ans);
    }

    return 0;
}

标程

// LA4329 Ping pong
// Rujia Liu
#include<cstdio>
#include<vector>
using namespace std;

//inline int lowbit(int x) { return x&(x^(x-1)); }
inline int lowbit(int x) { return x&-x; }

struct FenwickTree {
  int n;
  vector<int> C;

  void resize(int n) { this->n = n; C.resize(n); }
  void clear() { fill(C.begin(), C.end(), 0); }

  // ????A[1]+A[2]+...+A[x] (x<=n)
  int sum(int x) {
    int ret = 0;
    while(x > 0) {
      ret += C[x]; x -= lowbit(x);
    }
    return ret;
  }

  // A[x] += d (1<=x<=n)
  void add(int x, int d) {
    while(x <= n) {
      C[x] += d; x += lowbit(x);
    }
  }
};

const int maxn = 20000 + 5;
int n, a[maxn], c[maxn], d[maxn];
FenwickTree f;

int main() {
    freopen("la4329.txt","r",stdin);
  int T;
  scanf("%d", &T);
  while(T--) {
    scanf("%d", &n);
    int maxa = 0;
    for(int i = 1; i <= n; i++) { scanf("%d", &a[i]); maxa = max(maxa, a[i]); }
    f.resize(maxa);
    f.clear();
    for(int i = 1; i <= n; i++) {
      f.add(a[i], 1);
      c[i] = f.sum(a[i]-1);
    }
    f.clear();
    for(int i = n; i >= 1; i--) {
      f.add(a[i], 1);
      d[i] = f.sum(a[i]-1);
    }

    long long ans = 0;
    for(int i = 1; i <= n; i++)
      ans += (long long)c[i]*(n-i-d[i]) + (long long)(i-c[i]-1)*d[i];
    printf("%lld\n", ans);
  }
  return 0;
}