[HDU5592] ZYB's Premutation
[HDU5592] ZYB's Premutation
题目大意:一个由\([1,n]\)组成的数列,但不知道具体排列,但给出每个前缀的逆序对数目,让你还原排列
Solution
创造一颗\([1,n]\)的权值线段树,初始权值都为\(1\),我们从后往前离线处理,每次拿到一个前缀的逆序对数\(p[i]\),说明在\(i\)上的这个数字\(a_i\)前面有\(sum=p[i]-p[i-1]\)个数字比它大,所以\(a_i\)是\(a_1,\cdots ,a_i\)中第\(i -sum\)大的数字,查询后这个\(a_i\)对前面就没有用了,需要在权值线段树上删去
Code
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#define sc(x) scanf("%d", &x)
#define pf(x) printf("%d\n",x)
#define lson cur << 1
#define rson (cur << 1) | 1
const int N = 50005;
int num[N << 2], val[N << 2];
int p[N], ans[N];
int n;
void build(int cur, int l, int r){
num[cur] = r - l + 1;
if(l == r){
val[cur] = l;
return;
}else{
int mid = l + ((r - l) >> 1);
build(lson, l, mid);
build(rson, mid + 1, r);
}
}
int query(int cur, int l, int r, int k){
int mid = l + ((r - l) >> 1);
if(l == r){
// pf(val[cur]);
return val[cur];
}else{
if(num[lson] >= k){
return query(lson, l, mid, k);
}else{
return query(rson, mid + 1, r, k - num[lson]);
}
}
}
void update(int cur, int l, int r, int k){
num[cur]--;
if(l == r){
val[cur] = 0;
return;
}else{
int mid = l + ((r - l) >> 1);
if(k <= mid){
update(lson, l, mid, k);
}else{
update(rson, mid + 1, r, k);
}
}
}
int main(){
int t;
scanf("%d", &t);
while(t--){
scanf("%d", &n);
memset(num, 0, sizeof(num));
memset(val, 0, sizeof(val));
build(1, 1, n + 1);//l和r要和数组对应起来,不能建的时候是n+1,query和update时又变成了n
for(int i = 1, la; i <= n; ++i){
scanf("%d", &p[i]);
}
for(int i = n; i; --i){
ans[i] = query(1, 1, n + 1, i - p[i] + p[i - 1]);
update(1, 1, n + 1, ans[i]);
}
for(int i = 1; i < n; ++i){
printf("%d ", ans[i]);
}
printf("%d\n", ans[n]);
}
return 0;
}
Error
- \(l\)和\(r\)要和数组对应起来,不能建的时候是\(n+1\),\(query\)和\(update\)时又变成了\(n\)
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