POJ 2010 Moo University - Financial Aid

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <string.h>
 4 #include <queue>
 5 #include <algorithm>
 6 
 7 #define INF 2000000007
 8 using namespace std;
 9 
10 typedef pair<int,int> P;
11 
12 P cow[100007];
13 int r_aid[100007], l_aid[100007];
14 
15 bool cmp(P a, P b)
16 {
17     return a.second > b.second;
18 }
19 
20 priority_queue<P> ql, qr;//大根堆 按钱 从大到小排
21 int main()
22 {
23     int N, C;
24     int F;
25     int money = 0;
26     freopen("in.txt", "r", stdin);
27     cin >> N >> C >> F;
28     for (int i = 0; i < C; i++)
29     {
30         int score, aid;
31         scanf("%d%d", &score, &aid);
32         //cin >> score >> aid;
33         cow[i] = P(aid, score);
34     }
35     sort(cow, cow+C, cmp);//按score从大到小排
36     for (int i = 0; i < C; i++)//正扫一遍求左边
37     {
38         if (ql.size() < (N / 2) || i == 0)//避免N为1 的情况
39         {
40             ql.push(cow[i]);
41             money += cow[i].first;
42         }
43         else
44         {
45             l_aid[i] = money;//前N / 2的money情况
46             money += cow[i].first;
47             ql.push(cow[i]);
48             money -= ql.top().first;
49             ql.pop();
50         }
51         //l_aid[i] = money;//错误确实出在这里 这里前 N / 2个把自己加进去了
52     }
53     money = 0;
54     for (int i = C-1; i >= 0; i--)//反扫一遍求右边
55     {
56         if (C - i - 1 < N / 2 || i == C-1)
57         {
58             qr.push(cow[i]);
59             money += cow[i].first;
60         }
61         else
62         {
63             r_aid[i] = money;
64             money += cow[i].first;
65             qr.push(cow[i]);
66             money -= qr.top().first;
67             qr.pop();
68         }
69         //r_aid[i] = money;已经重新算过 money了 出现了错误
70     }
71     for (int i = 0+N / 2; i <= C - 1 - N / 2; i++)//从左往右找到第一个 小于l_aid[i]+r_aid[i]的
72     {
73         if(l_aid[i] + r_aid[i] + cow[i].first <= F)
74         {
75             cout << cow[i].second << endl;
76             return 0;
77         }
78     }
79     cout << -1 << endl;
80     return 0;
81 }