【POJ2182】Lost Cows

#【POJ2182】Lost Cows

##题面

vjudge

##题解

从后往前做

每扫到一个点$i$以及比前面小的有$a[i]$个数

就是查询当前的第$a[i]+1$小

然后查询完将这个数删掉

两个操作可以用平衡树实现

但是我比较懒用了$01trie$

据说暴力也可以过

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm> 
using namespace std;

inline int gi() {
    register int data = 0, w = 1;
    register char ch = 0;
    while (ch != '-' && (ch > '9' || ch < '0')) ch = getchar();
    if (ch == '-') w = -1 , ch = getchar();
    while (ch >= '0' && ch <= '9') data = data * 10 + (ch ^ 48), ch = getchar();
    return w * data;
}
const int MAX_N = 80005;
struct Trie { int ch[2], size; } t[MAX_N * 18]; int tot; 
void insert(int v) { 
	int o = 1; t[o].size++; 
	for (int i = 17; i >= 0; i--) { 
		int c = v >> i & 1; 
		if (!t[o].ch[c]) t[o].ch[c] = ++tot; 
		t[o = t[o].ch[c]].size++; 
	} 
} 
void erase(int v) {
	int o = 1; t[o].size--; 
	for (int i = 17; i >= 0; i--) { 
		int c = v >> i & 1; 
		t[o = t[o].ch[c]].size--; 
	} 
} 
int Kth(int k) {
	int o = 1, res = 0; 
	for (int i = 17; i >= 0; i--) {
		int sz = t[t[o].ch[0]].size; 
		if (k <= sz) o = t[o].ch[0]; 
		else k -= sz, res |= 1 << i, o = t[o].ch[1]; 
	} 
	return res; 
}
int N, a[MAX_N], ans[MAX_N]; 
int main () {
	N = gi(); for (int i = 1; i < N; i++) a[i + 1] = gi(); ++tot; 
	for (int i = 1; i <= N; i++) insert(i); 
	for (int i = N; i >= 1; i--) { 
		ans[i] = Kth(a[i] + 1); 
		erase(ans[i]);
	}
	for (int i = 1; i <= N; i++) printf("%d
", ans[i]); 
    return 0; 
}