POJ 2010 Moo University - Financial Aid （优先队列/二分答案）

题目大意：

从c（<=1e5）个物品中选出n（奇数 <= 19,999）个物品，使得这n个物品的价值中位数尽量高，且总价格不超过f (<= 2,000,000,000)。

做法1：

• 把物品按价值排序
• half表示n/2向下取整
• low[i]表示前i - 1个物品中价格最小的half个物品的价格和
• hei[i]表示i之后的物品中价格最小的half个物品的价格和
• low和hei可用优先队列维护
• 枚举中间物品即可

代码

NKWBTB	Accepted	2592KB	204MS	C++	1154

#include 
#include 
#include 
#include 
#include 

using namespace std;

const int maxc = 1e5 + 10;

pair<int, int>p[maxc];

long long low[maxc], hei[maxc];

int main()
{
    int n, c, f;
    scanf("%d%d%d", &n, &c, &f);
    int half = n / 2;
    for (int i = 0 ; i < c ; i++)
        scanf("%d%d", &p[i].first, &p[i].second);
    sort(p, p + c);
    {
        long long sum = 0;
        priority_queue<long long>q;
        for (int i = 0 ; i < half ; i++)
        {
            q.push(p[i].second);
            sum += p[i].second;
            low[i] = 0x7fffffff;
        }
        for (int i = half ; i < c ; i++)
        {
            low[i] = sum;
            q.push(p[i].second);
            sum += p[i].second - q.top();
            q.pop();
        }
    }
    {
        long long sum = 0;
        priority_queue<long long>q;
        for (int i = c - 1 ; i >= c - half ; i--)
        {
            q.push(p[i].second);
            sum += p[i].second;
            hei[i] = 0x7fffffff;
        }
        for (int i = c - half - 1 ; i >= 0 ; i--)
        {
            hei[i] = sum;
            q.push(p[i].second);
            sum += p[i].second - q.top();
            q.pop();
        }
    }
    int ans = -1;
    for (int i = c - 1 ; i >= 0 ; i--)
        if (hei[i] + low[i] + p[i].second <= f)
        {
            ans = p[i].first;
            break;
        }
    printf("%d\n", ans);
    return 0;
}

做法2:

• 保存两组信息，一组物品按价值排序，另一组按价格排序
• 二分中间物品的价格
• 检查答案是否合法

代码

NKWBTB	Accepted	2512KB	235MS	C++	1178

#include 
#include 
#include 
#include 

using namespace std;

const int maxn = 1e5 + 10;

struct stu
{
    int first, second, id;
}p[maxn], s[maxn];

bool cmp1(const stu &a, const stu &b){return a.first < b.first;}
bool cmp2(const stu &a, const stu &b){return a.second < b.second;}

int main()
{
    int n, c ,f;
    scanf("%d%d%d", &n, &c, &f);
    int half = n / 2;
    for (int i = 0 ; i < c ; i++)
        scanf("%d%d", &p[i].first, &p[i].second);
    sort(p, p + c, cmp1);
    for (int i = 0 ; i < c ; i++)
        p[i].id = i;
    memcpy(s, p, sizeof(stu) * c);
    sort(p, p + c, cmp2);
    int l = 0, r = c, ans = -1;
    while (l < r)
    {
        int m = (l + r) / 2;
        int lc = 0, rc = 0, sum = s[m].second;
        for (int i = 0 ; i < c ; i++)
        {
            if (lc < half && p[i].id < s[m].id && sum + p[i].second <= f)
            {
                sum += p[i].second;
                lc++;
            }else if (rc < half && p[i].id > s[m].id && sum + p[i].second <= f)
            {
                sum += p[i].second;
                rc++;
            }
        }
        if (lc < half && rc < half)
        {
            ans = -1;
            break;
        }
        else if (lc < half)
            l = m + 1;
        else if (rc < half)
            r = m;
        else{
            ans = s[m].first;
            l = m + 1;
        }
    }

    printf("%d\n", ans);
    return 0;
}