NWERC 2019 A

A. Average Rank

The National Weekly Escape Room Challenge (NWERC) is a long-running competition held in Eindhoven. Every week a new escape room is presented, and anyone who completes it in their first attempt gains one point.

At the end of each week, competitors are ranked by the total number of points accumulated so far, highest first. In the case of a tie, they share the same rank. In other words, the rank of a competitor is one more than the number of people with a strictly larger number of points.

In total there have been \(n\) participants in the contest, and the contest has been going for w weeks. For each week you are given a list of the competitors that gained a point that week. Your task is to calculate the average rank during the w-week competition for each competitor.
The figure illustrates the score progression in the third sample.

Input
The input consists of:

One line with two integers \(n\) and \(w\) (\(1≤n,w≤3⋅10^5\)), the number of competitors and the number of weeks. The competitors are numbered from \(1\) to \(n\).
\(w\) lines (one for each week), each containing an integer \(k\) (\(0≤k≤n\)) followed by \(k\) distinct integers \(c_1,…,c_k\) (\(1≤c_i≤n\) for all \(i\)), indicating that the \(k\) competitors \(c_1,…,c_k\) each gained a point that week.
The total number of points awarded is at most 1 million.

Output
Output \(n\) lines, the ith of which contains the average rank of the \(ith\) competitor during the \(w\)-week competition. Your answers should have an absolute or relative error of at most \(10^{−6}\).

題解

這題說實話思惟量很大。

其實咱們要先想到，操做總數小於一百萬，也就是說，對於一個參賽者分數\(+1\), 受到修改的，只有和他同分的人以及他本身，修改一次，與他同分的人加一，他本身的排名也會隨之變化。

而後咱們考慮每一個人的平均排名，實際上就是每週的排名之和。

假設咱們暴力處理的話， 效率就是\(O(n*w)\).

考慮優化這個過程。

類比修改線段樹的\(lazy_tag\), 咱們對於每次\(+1\)修改的，利用差分的思想，到用到的時候再算進去。

說實話有點難想。

Code

#include<bits/stdc++.h>
#define _(d) while(d(isdigit(ch=getchar())))
#define rep(i,a,b) for(int i=a;i<=b;i++)
template<class T>void g(T&t){T x,f=1;char ch;_(!)ch=='-'?f=-1:f;x=ch-48;_()x=x*10+ch-48;t=f*x;}
using namespace std;
const int N = 1e6 + 4;
typedef long long ll;
ll n, w, now[N], pre[N], cnt[N], p[N], sum[N];
int main(){
	g(n), g(w);
	cnt[1] = n;
	rep( i, 0, n ) sum[i] = w, p[i] = 1;
	rep( i, 1, w ){
		int Len; g(Len);
		while( Len-- ){
			int x; g(x);
			sum[ x ] += now[ p[x] ] - pre[ x ];
			now[ p[x] ] += ( w - i + 1 );
			cnt[ p[x] ] --;
			p[ x ] ++;
			sum[ x ] -= cnt[ p[x] ] * ( w - i + 1 );
			pre[ x ] = now[ p[x] ];
			cnt[ p[x] ]++;
		}
	}
	rep( i, 1, n ){
		sum[i] += now[ p[i] ] - pre[i];
		printf("%.11lf\n",1.0 * sum[i] / ( 1.0 * w ));
	}
	return 0;
}