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Educational Codeforces Round 131 D - Permutation Restoration

作者:互联网

对于每个bi,可以求出ai属于 [(i / (bi + 1)) + 1 , i / bi]
然后就是贪心,参考了yyg的写法
依次枚举1~n,枚举到第i个时,把左边界为i的都放进优先队列,此时优先队列中所有元素的左边界<=i,取出一个右边界最小的即可
因为保证有解,所以取出的元素右边界一定>=i(反证:如果右边界<i,那么肯定属于i之前的点,而i之前的点都是通过贪心得到的,所以无解)
#include<bits/stdc++.h>
using namespace std;

#define fr first
#define se second
#define et0 exit(0);
#define rep(i, a, b) for(int i = (int)(a); i <= (int)(b); i ++)
#define rrep(i, a, b) for(int i = (int)(a); i >= (int)(b); i --)

typedef long long LL;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
typedef unsigned long long ULL;

const int INF = 0X3f3f3f3f, N = 5e5 + 10, MOD = 1e9 + 7;
const double eps = 1e-7;

int a[N];

vector<PII> g[N];

priority_queue<PII, vector<PII>, greater<PII> > q;

void work() {
	int n;
	cin >> n;
	rep(i, 1, n) g[i].clear();
	rep(i, 1, n) {
		int x, l, r;
		cin >> x;
		l = (i / (x + 1)) + 1;
		if (x == 0) r = n;
		else r = i / x;
		g[l].push_back({r, i});
	}
	rep(i, 1, n) {
		rep(j, 0, g[i].size() - 1) q.push(g[i][j]);
		a[q.top().se] = i;
		q.pop();
	}
	rep(i, 1, n) cout << a[i] << " ";
	cout << endl;
}

signed main() {
	int test;
	cin >> test;

	while (test--) {
		work();
	}

	return 0;
}

标签:Educational,int,rep,typedef,long,Codeforces,bi,131,define
来源: https://www.cnblogs.com/xhy666/p/16462348.html