题目传送门：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 bi and pi (1 ≤ bi, pi ≤ 100 000) — the beauty and the cost of the i-th fountain, and then a letter "C" or "D", describing in which type of money is the cost of fountain i: in coins or in diamonds, respectively.
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.
题目大意：
有 N 个水池，每个水池有 观赏值 和 花费（金币或者钻石）；Arkady有 C 个金币 D 个钻石，他想建两个水池，使得观赏值最高。
解题思路：
树状数组维护金币和钻石花费范围内所能得到的最大值，每次输入都比较三种可能性一个金币的一个钻石的，两个金币的，两个钻石的。
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 ", ans); 58 return 0; 59 }