[C++]Knights of a Polygonal Table（骑士的多角桌）

+ 问题描述

　　+ Link: https://codeforces.com/contest/994/problem/B

　　+ 描述

Unlike Knights of a Round Table, Knights of a Polygonal Table deprived of nobility and happy to kill each other. But each knight has some power and a knight can kill another knight if and only if his power is greater than the power of victim. However, even such a knight will torment his conscience, so he can kill no more than kk other knights. Also, each knight has some number of coins. After a kill, a knight can pick up all victim's coins.

Now each knight ponders: how many coins he can have if only he kills other knights?

You should answer this question for each knight.

　　+ 大意

　　　　n个骑士, 每个骑士有战力p, 钱c, 每个骑士可以抢战力比他低的钱, 每个骑士最多抢k次, 对每个骑士求出最大钱数。

+ 输入

The first line contains two integers nn and kk (1≤n≤10^5,0≤k≤min(n−1,10))(1≤n≤10^5,0≤k≤min(n−1,10)) — the number of knights and the number kk from the statement.

The second line contains nn integers p1,p2,…,pn (1≤pi≤10^9)(1≤pi≤10^9) — powers of the knights. All pipi are distinct.

The third line contains nn integers c1,c2,…,cn (0≤ci≤10^9)(0≤ci≤10^9) — the number of coins each knight has.

+ 输出

Print $\mathrm{nn integers\; —\; the\; maximum\; number\; of\; coins\; each\; knight\; can\; have\; it\; only\; he\; kills\; other\; knights.}$

+ 网友思路

　　1 贪心 +(优先队列)

　　2 线段树

　　n ...

+ 代码

【程序结果：用例未完全通过，本博文仅为暂存代码之目的】

/*
	B. Knights of a Polygonal Table
		url:http://codeforces.com/problemset/problem/994/B
	思路：
		step0.对骑士进行编号 
		step1.按照权力值进行顺序排序
		step2.分别计算骑士所能杀死的其他骑士及其对应的金币值
		step3.按照编号重新恢复原始排序. 
*/
//#include<bits/stdc++.h> 
#include<iostream>
#include<algorithm>
using namespace std;

typedef struct Knight{
	int id;
	int power;
	int coins;
	int expected_coins; 
}Knight;

bool compare_id(Knight a, Knight b){
	if(a.id>b.id)
		return false;
	else return true; 
}

bool compare_power(Knight a, Knight b){
	if(a.power>b.power)
		return false;
	else return true; 
}

int main(){
	int n,k,id = 0;
	Knight *persons;
	scanf("%d %d", &n, &k);
	persons = (Knight*)malloc(n*sizeof(Knight));
	//printf("
%d %d", k, n);
	for(int i=0;i<n;i++){
		persons[i].id = ++id;
		scanf("%d", &(persons[i].power));
	}
	for(int i=0;i<n;i++){
		scanf("%d", &(persons[i].coins));
	}
	
	sort(persons, persons+n, compare_power);//顺序排序：<algorithm> 
	
	for(int i=0;i<n;i++){//分别计算各骑士能够杀死的人和可以获得的金币最大值 
		persons[i].expected_coins = persons[i].coins;
		for(int j=1;j<=k;j++){
			if(i - j >= 0){
				persons[i].expected_coins += persons[i - j].coins;	
			}
		}
	}
	
	sort(persons, persons+n, compare_id);//依照id恢复原始排序 
	
	for(int i=0;i<n;i++){
		printf("%d%s", persons[i].expected_coins, ((i + 1) != n ?" ":"
"));//test	
		//printf("%d %d %d
", persons[i].power, persons[i].coins, persons[i].expected_coins);//test
	}
	
	return 0;
}