POJ2454 Jersey Politics

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.

先贪心的把数组从大到小排序，分3组

最小的一组无论是否合格，都不管

接下来随机交换第1组和第2组的值，直到有解

这应该属于随机算fa♂(即通过足够长的时间搞出答案才停)

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<algorithm>
 #include<ctime>
 using namespace std;
 struct ZYYS
 {
   int x,id;
 }a[];
 int k,sum1,sum2;
 bool cmp(ZYYS a,ZYYS b)
 {
   return a.x>b.x;
 }
 int main()
 {int i,j;
   cin>>k;
   for (i=;i<=*k;i++)
     {
       scanf("%d",&a[i].x);
       a[i].id=i;
     }
   sort(a+,a+*k+,cmp);
   for (i=;i<=k;i++)
     sum1+=a[i].x,sum2+=a[i+k].x;
   srand(time());
   while (!(sum1>*k&&sum2>*k))
     {
       int x1=rand()%k+,x2=rand()%k++k;
       sum1=sum1-a[x1].x+a[x2].x;
       sum2=sum2-a[x2].x+a[x1].x;
       swap(a[x1],a[x2]);
     }
   for (i=;i<=*k;i++)
     printf("%d\n",a[i].id);
 }