ZOJ 3512 Financial Fraud (左偏树)

Posted 2020-09-07 dwtfukgv

题意：给定一个序列，求另一个不递减序列，使得Abs(bi - ai) 和最小。

析：首先是在每个相同的区间中，中位数是最优的，然后由于要合并，和维护中位数，所以我们选用左偏树来维护，当然也可以用划分树来做。

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>
#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 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 double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 50000 + 10;
const int mod = 1e6 + 10;
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 int Abs(int x){ return x > 0 ? x : -x; }
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}
struct Node{ int key, l, r, d, fa; };
Node tr[maxn];
int iroot(int i){
  if(i == -1)  return -1;
  while(tr[i].fa != -1)  i = tr[i].fa;
  return i;
}

int Merge(int rx, int ry){
  if(rx == -1)  return ry;
  if(ry == -1)  return rx;
  if(tr[rx].key < tr[ry].key)  swap(rx, ry);
  int r = Merge(tr[rx].r, ry);
  tr[rx].r = r;  tr[r].fa = rx;
  if(tr[tr[rx].l].d < tr[r].d)  swap(tr[rx].l, tr[rx].r);
  if(tr[rx].r == -1)  tr[rx].d = 0;
  else  tr[rx].d = tr[tr[rx].r].d + 1;
  return rx;
}

int del(int i){
  if(i == -1) return -1;
  int l = tr[i].l, r = tr[i].r, y = tr[i].fa, x;
  tr[i].l = tr[i].r = tr[i].fa = -1;
  tr[x = Merge(l, r)].fa = y;
  if(y != -1 && tr[y].l == i)  tr[y].l = x;
  else if(y != -1 && tr[y].r == i)  tr[y].r = x;
  for( ; y != -1; x = y, y = tr[y].fa){
    if(tr[tr[y].l].d < tr[tr[y].r].d)  swap(tr[y].l, tr[y].r);
    if(tr[y].d == tr[tr[y].r].d + 1)  break;
    tr[y].d = tr[tr[y].r].d + 1;
  }
  if(x != -1)  return iroot(x);
  return iroot(y);
}

int top(int i){ return tr[i].key; }
int pop(int &i){
  Node out = tr[i];
  int l = tr[i].l, r = tr[i].r;
  tr[i].l = tr[i].r = tr[i].fa = -1;
  tr[l].fa = tr[r].fa = -1;
  i = Merge(l, r);
  return out.key;
}

int a[maxn];
void init(){
  for(int i = 0; i < n; ++i){
    scanf("%d", a+i);
    tr[i].key = a[i];
    tr[i].l = tr[i].r = tr[i].fa = -1;
    tr[i].d = 0;
  }
}
int tree[maxn], sz[maxn], cnt[maxn];

void solve(){
  int m = -1;
  for(int i = 0; i < n; ++i){
    tree[++m] = i;
    sz[m] = cnt[m] = 1;
    while(m > 0 && top(tree[m-1]) >= top(tree[m])){
      tree[m-1] = Merge(tree[m-1], tree[m]);
      sz[m-1] += sz[m];
      cnt[m-1] += cnt[m];
      --m;
      while(cnt[m] > (sz[m]+1) / 2){
        pop(tree[m]);
        --cnt[m];
      }
    }
  }
  LL ans = 0;
  int k = 0;
  for(int i = 0; i <= m; ++i){
    int t = top(tree[i]);
    for(int j = 0; j < sz[i]; ++j, ++k)
      ans += Abs(t - a[k]);
  }
  printf("%lld\n", ans);
}

int main(){
  while(scanf("%d", &n) == 1 && n){
    init();
    solve();
  }
  return 0;
}