題目傳送門：http://codeforces.com/contest/799/problem/C

C. Fountains

Arkady plays Gardenscapes a lot. Arkady wants to build two new fountains. There are n available fountains, for each fountain its beauty and cost are known. There are two types of money in the game: coins and diamonds, so each fountain cost can be either in coins or diamonds. No money changes between the types are allowed.

Help Arkady to find two fountains with maximum total beauty so that he can buy both at the same time.

Input

The first line contains three integers n, c and d (2?≤?n?≤?100?000, 0?≤?c,?d?≤?100?000) — the number of fountains, the number of coins and diamonds Arkady has.

The next n lines describe fountains. Each of these lines contain two integers b

Output

Print the maximum total beauty of exactly two fountains Arkady can build. If he can‘t build two fountains, print 0.

3 7 6
10 8 C
4 3 C
5 6 D

9

2 4 5
2 5 C
2 1 D

0

3 10 10
5 5 C
5 5 C
10 11 D

10

Note

In the first example Arkady should build the second fountain with beauty 4, which costs 3 coins. The first fountain he can‘t build because he don‘t have enough coins. Also Arkady should build the third fountain with beauty 5 which costs 6 diamonds. Thus the total beauty of built fountains is 9.

In the second example there are two fountains, but Arkady can‘t build both of them, because he needs 5 coins for the first fountain, and Arkady has only 4 coins.

題目大意：

有 N 個水池，每個水池有 觀賞值 和 花費（金幣或者鉆石）；Arkady有 C 個金幣 D 個鉆石，他想建兩個水池，使得觀賞值最高。

解題思路：

樹狀數組維護金幣和鉆石花費範圍內所能得到的最大值，每次輸入都比較三種可能性一個金幣的一個鉆石的，兩個金幣的，兩個鉆石的。

AC code：

 1 #include <bits/stdc++.h>
 2 #define INF 0x3f3f3f3f
 3 #define ll long long int
 4 using namespace std;
 5 
 6 const int MAXN = 1e5+10;
 7 
 8 int t_1[MAXN], t_2[MAXN];
 9 int N, num_c, num_d;
10 
11 int lowbit(int x)
12 {
13     return x&(-x);
14 }
15 void add(int no, int st, int value)
16 {
17     for(int i = st; i <= MAXN; i+=lowbit(i))
18     {
19         if(no) t_1[i] = max(t_1[i], value);
20         else t_2[i] = max(t_2[i], value);
21     }
22 }
23 
24 int query(int no, int st)
25 {
26     int res = 0;
27     for(int i = st; i > 0; i-=lowbit(i))
28     {
29         if(no) res = max(res, t_1[i]);
30         else res = max(res, t_2[i]);
31     }
32     return res;
33 }
34 int main()
35 {
36     char str[3];
37     int ans_max = 0, a, b, ans = 0;
38     scanf("%d%d%d", &N, &num_c, &num_d);
39     for(int i = 1; i <= N; i++)
40     {
41         scanf("%d%d", &a, &b);
42         scanf("%s", str);
43         if(str[0] == ‘C‘){
44             ans_max = query(0, num_d);
45             if(b > num_c) continue;
46             ans_max = max(ans_max, query(1, num_c-b));
47             add(1, b, a);
48         }
49         else{
50             ans_max = query(1, num_c);
51             if(b > num_d) continue;
52             ans_max = max(ans_max, query(0, num_d-b));
53             add(0, b, a);
54         }
55         if(ans_max) ans = max(ans, ans_max+a);
56     }
57     printf("%d\n", ans);
58     return 0;
59 }

Playrix Codescapes Cup (Codeforces Round #413, rated, Div. 1 + Div. 2) C. Fountains 【樹狀數組維護區間最大值】