POJ2454:Jersey Politics(贪心+随机化) java程序员

Jersey Politics

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4231		Accepted: 1022		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.

Source

USACO 2005 February Gold

MYCode:

#include<iostream>
#include<cstring>
#include<cstdio>
#include<ctime>
#include<algorithm>
using namespace std;
#define MAX 200
struct node
{
    int v;
    int id;
}a[MAX];
bool cmp(node a,node b)
{
    return a.v>b.v;
}
int main()
{
    int k;
    while(scanf("%d",&k)!=EOF)
    {
        int i;
        for(i=1;i<=3*k;i++)
        {
            scanf("%d",&a[i].v);
            a[i].id=i;
        }
        sort(a+1,a+3*k+1,cmp);
        int s1=0,s2=0;
        for(i=1;i<=k;i++)
        {
            s1+=a[i].v;
            s2+=a[i+k].v;
        }
        srand((int)(time(0)));
        while(true)
        {
            if(s1>500*k && s2>500*k)
            break;
            int id1=rand()%k+1;
            int id2=rand()%k+1+k;
            if(a[id1].v <a[id2].v&&s1<s2||a[id1].v>a[id2].v&&s1>s2)
            {
                s1+=a[id2].v-a[id1].v;
                s2+=a[id1].v-a[id2].v;
                node tp;
                tp=a[id2];
                a[id2]=a[id1];
                a[id1]=tp;
            }
        }
        for(i=1;i<=k;i++)
        {
            printf("%d\n",a[i].id);
        }
        for(i=k+1;i<=2*k;i++)
        {
            printf("%d\n",a[i].id);
        }
        for(i=2*k+1;i<=3*k;i++)
        {
            printf("%d\n",a[i].id);
        }
    }
}
//16MS

贪心+随机化