【数学】Jersey Politics

                                                        Jersey Politics

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5592		Accepted: 1413		Special Judge

Description

In the newest census of Jersey Cows and Holstein Cows, Wisconsin cows have earned three stalls in the Barn of Representatives. The Jersey Cows currently control the state's redistricting committee. They want to partition the state into three equally sized voting districts such that the Jersey Cows are guaranteed to win elections in at least two of the districts.

Wisconsin has 3*K (1 <= K <= 60) cities of 1,000 cows, numbered 1..3*K, each with a known number (range: 0..1,000) of Jersey Cows. Find a way to partition the state into three districts, each with K cities, such that the Jersey Cows have the majority percentage in at least two of districts.

All supplied input datasets are solvable.

Input

* Line 1: A single integer, K

* Lines 2..3*K+1: One integer per line, the number of cows in each city that are Jersey Cows. Line i+1 contains city i's cow census.

Output

* Lines 1..K: K lines that are the city numbers in district one, one per line

* Lines K+1..2K: K lines that are the city numbers in district two, one per line

* Lines 2K+1..3K: K lines that are the city numbers in district three, one per line

Sample Input

2
510
500
500
670
400
310

Sample Output

1
2
3
6
5
4

Hint

Other solutions might be possible. Note that "2 3" would NOT be a district won by the Jerseys, as they would be exactly half of the cows.

 

试题分析：

    我们可以想到把数组排一个序，然后取前2k个，分到S1、S2中去

    设S=(S1，S2)

    那么我们持续随机排序S，最终在满足条件时输出答案

  

    上面的策略固然能解决这个问题，但是随即排序S时间浪费太大

    那么我们要怎样呢？

    我们可以随机交换S1，S2中的元素，然后直到满足条件就可以了

 

代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<stdlib.h>
#include<time.h>
using namespace std;
inline int read(){
	int x=0,f=1;char c=getchar();
	for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
	for(;isdigit(c);c=getchar()) x=x*10+c-'0';
	return x*f;
}
struct data{
	int num,k;
}a[200],s1[61],s2[61],s3[61];//S1 S2满足条件
bool cmp(data a,data b){
	return a.num>b.num;
}
int N;
bool judge(){
	int sum=0,sum2=0;
	for(int i=1;i<=N/3;i++) sum+=s1[i].num;//S1 sum
	for(int i=1;i<=N/3;i++) sum2+=s2[i].num;
	if(sum2>500*(N/3)&&sum>500*(N/3)) return true;
	return false;
}
void change(){
	//srand((unsigned int)time(0));加上这个POJ会给你显示RE
	int tmp1=rand()%(N/3)+1;
	int tmp2=rand()%(N/3)+1;
	swap(s1[tmp1].num,s2[tmp2].num);
	swap(s1[tmp1].k,s2[tmp2].k);
}
void print(){
	for(int i=1;i<=N/3;i++) cout<<s1[i].k<<endl;
	for(int i=1;i<=N/3;i++) cout<<s2[i].k<<endl;
	for(int i=1;i<=N/3;i++) cout<<s3[i].k<<endl;
}
int main(){
	N=read(),N*=3;
	for(int i=1;i<=N;i++) a[i].num=read(),a[i].k=i;
	sort(a+1,a+N+1,cmp);

	for(int i=1;i<=N/3;i++) //划分前k大到S1
	    s1[i].num=a[i].num,s1[i].k=a[i].k;
	for(int i=N/3+1;i<=2*(N/3);i++) //划分前K+1~前2K大到S2
	    s2[i-N/3].num=a[i].num,s2[i-N/3].k=a[i].k;
	for(int i=2*(N/3)+1;i<=N;i++)
	    s3[i-2*(N/3)].num=a[i].num,s3[i-2*(N/3)].k=a[i].k;

    while(!judge()) change();
    print();
}