P1966 火柴排队
P1966 火柴排队
树状数组的简单练习题目
题目要求最小化 \(\sum(a_i-b_i)^2\) 可以转化成 a数组中第K大的数要和b数组中第K大的数在同一个位置,所以先做一手离散化
然后令\(pos[brr[i]] = i\) pos数组中存储的是 b数组值所对应的位置
\(new[i] = pos[arr[i]]\) new数组中存储的是 a数组值在b数组值中对应的位置
\(对new[i]求逆序对\)
假设 $b[i] = $ {\(5,3,4,2,1\)}
有 \(a[i] =\) {\(4,3,5,2,1\)}
\(pos[i]=\) {\(5,4,2,3,1\)}
\(new[i]=\) {\(3,2,1,4,5\)}
最后就是要把这个顺序给他升序一下 使得\(new[i] = i\) 说明此时a数组完全等于b数组
这个操作次数就等于new序列的逆序对个数
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <ctime>
#include <iostream>
#define lowbit(a) a&(-a)
using namespace std;
const int maxn = 1e5+5;
const int mod = 1e8-3;
int N;
long long arr[500005], brr[500005], Trarr[500005], sorting[2][500005], Newarr[500005], pos[500005];
;
void add(int x, long long val){
while(x <= N){
Trarr[x] += val;
x += lowbit(x);
}
return;
}
long long getsum(int x){
long long ans = 0;
while(x){
ans += Trarr[x];
x -= lowbit(x);
}
return ans;
}
signed main(){
clock_t c1 = clock();
#ifdef LOCAL
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif
scanf("%d", &N);
for(int i = 1; i <= N; i++) scanf("%lld", &arr[i]);
for(int i = 1; i <= N; i++){
scanf("%lld", &brr[i]);
sorting[0][i] = arr[i];
sorting[1][i] = brr[i];
}
sort(sorting[0]+1, sorting[0]+1+N);
sort(sorting[1]+1, sorting[1]+1+N);
int lena = unique(sorting[0]+1, sorting[0]+1+N) - sorting[0] - 1;
int lenb = unique(sorting[1]+1, sorting[1]+1+N) - sorting[1] - 1;
for(int i = 1; i <= N; i++){
arr[i] = lower_bound(sorting[0]+1, sorting[0]+1+lena, arr[i]) - sorting[0];
brr[i] = lower_bound(sorting[1]+1, sorting[1]+1+lenb, brr[i]) - sorting[1];
pos[brr[i]] = i;
}
for(int i = 1; i <= N; i++) Newarr[i] = pos[arr[i]];
long long ans = 0;
for(long long i = 1; i <= N; i++){
add(Newarr[i], 1);
ans = (ans + (i-getsum(Newarr[i]))) % mod;
}
printf("%lld", ans);
end:
cerr << "Time Used: " << clock() - c1 << "ms" << endl;
return 0;
}