HDU 5792 World is Exploding (离散化+树状数组)

Posted 2020-10-05 dwtfukgv

题意：给定 n 个数，让你数出 a < b && c < d && a != b != c != d  && Aa < Ab && Ac > Ad。

析：首先，给的数太大了，先要进行离散化处理，然后先算出Aa < Ab 和  Ac > Ad。这可以用树状数组解决，一个正向的，一个反向，同时再求出四种数，然后减去，就好了。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
//#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e15;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 5e4 + 100;
const int mod = 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r >= 0 && r < n && c >= 0 && c < m;
}

vector<int> v;

int a[maxn];
int sum[2][maxn<<1];

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

void add(int pos, int x){
  while(x <= n){
    ++sum[pos][x];
    x += lowbit(x);
  }
}

int query(int pos, int x){
  int ans = 0;
  while(x){
    ans += sum[pos][x];
    x -= lowbit(x);
  }
  return ans;
}

int getPos(int x){
  return lower_bound(v.begin(), v.end(), x) - v.begin();
}

int b[maxn], c[maxn], d[maxn], e[maxn];

int main(){
  while(scanf("%d", &n) == 1){
    v.cl; ms(sum, 0);
    v.pb(-1);
    for(int i = 0; i < n; ++i){
      scanf("%d", a+i);
      v.pb(a[i]);
    }
    sort(v.begin(), v.end());
    v.erase(unique(v.begin(), v.end()), v.end());

    LL ans = 0;
    LL tmp1 = 0, tmp2 = 0;
    LL tmp3 = 0, tmp4 = 0, tmp5 = 0;
    for(int i = 0; i < n; ++i){
      int pos = getPos(a[i]);
      b[i] = query(0, pos-1);
      tmp1 += b[i]; // min
      c[i] = i - query(0, pos);
      tmp3 += (LL)b[i] * c[i];
      add(0, pos);
    }

    for(int i = n-1; i >= 0; --i){
      int pos = getPos(a[i]);
      d[i] = query(1, pos-1);
      tmp2 += d[i];  //right min
      e[i] = (n-i-1) - query(1, pos);
      tmp4 += (LL)d[i] * e[i];
      //tmp4 += (n-i-1) - query(1, pos); // right max
      add(1, pos);
    }

    LL tmp6 = 0;
    for(int i = 0; i < n; ++i){
      tmp5 += (LL)c[i] * e[i];
      tmp6 += (LL)b[i] * d[i];
    }

    ans = tmp1 * tmp2 - tmp3 - tmp4 - tmp5 - tmp6;
    printf("%I64d\n", ans);

  }
  return 0;
}