poj_2454 Jersey Politics（贪心+随机）

Jersey Politics

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5596		Accepted: 1416		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

Sample Output

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，对数组排序后，其中一个最小的划分一定是数组前面的k个元素，另外两个划分则在后面的2*k个元素中。

然后先假设两个划分刚好是k+1~2*k和2*k~3*k，这种划分不一定满足题目要求的两个划分各自的元素和都大于500*k，

所以需要对两个划分进行调整，可以用随机算法随机从两个划分中各抽取一个元素进行交换，直到满足题意。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <stack>
#include <bitset>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <algorithm>
#define FOP freopen("data.txt","r",stdin)
#define FOP2 freopen("data1.txt","w",stdout)
#define inf 0x3f3f3f3f
#define maxn 200
#define mod 1000000007
#define PI acos(-1.0)
#define LL long long
using namespace std;

struct Node
{
    int x, pos;
}a[maxn];

bool cmp(Node a, Node b)
{
    return a.x < b.x;
}

int k;
Node x1[maxn], x2[maxn];

bool judge()
{
    int s1 = 0, s2 = 0;
    for(int i = 1; i <= k; i++) s1 += x1[i].x, s2 += x2[i].x;
    if(s1 > 500*k && s2 > 500*k) return true;
    else return false;
}

void solve()
{
    srand((unsigned)time(NULL));
    while(1)
    {
        if(judge()) break;
        int i = rand()%k+1;
        int j = rand()%k+1;
        swap(x1[i], x2[j]);
    }
}

int main()
{
    while(~scanf("%d", &k))
    {
        for(int i = 1; i <= 3*k; i++) scanf("%d", &a[i].x), a[i].pos = i;
        sort(a+1, a+3*k+1, cmp);
        for(int i = 1; i <= k; i++) x1[i] = a[i+k], x2[i] = a[i+2*k];
        solve();
        for(int i = 1; i <= k; i++) printf("%d\n", a[i].pos);
        for(int i = 1; i <= k; i++) printf("%d\n", x1[i].pos);
        for(int i = 1; i <= k; i++) printf("%d\n", x2[i].pos);
    }
    return 0;
}

poj_2454 Jersey Politics（贪心+随机）

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