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.
Example
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.
1 #include <iostream> 2 #include <string> 3 #include <algorithm> 4 #include <cstring> 5 #include <cstdio> 6 #include <cmath> 7 #include <queue> 8 #include <set> 9 #include <map> 10 #include <list> 11 #include <stack> 12 #define mp make_pair 13 typedef long long ll; 14 typedef unsigned long long ull; 15 const int MAX=1e5; 16 const int INF=1e9+5; 17 const double M=4e18; 18 using namespace std; 19 const int MOD=1e9+7; 20 typedef pair<int,int> pii; 21 const double eps=0.000000001; 22 int C[MAX+5],D[MAX+5]; 23 void add(int *tree,int k,int num) 24 { 25 while(k<=MAX) 26 { 27 tree[k]=max(tree[k],num); 28 k+=k&(-k); 29 } 30 } 31 int read(int *tree,int k) 32 { 33 int re=0; 34 while(k) 35 { 36 re=max(re,tree[k]); 37 k-=k&(-k); 38 } 39 return re; 40 } 41 int n,c,d,b,p; 42 int an,tem; 43 char opt[10]; 44 int main() 45 { 46 scanf("%d%d%d",&n,&c,&d); 47 while(n--) 48 { 49 scanf("%d%d%s",&b,&p,opt); 50 tem=0; 51 if(opt[0]=='C') 52 { 53 if(p>c) 54 continue; 55 tem=read(D,d); 56 tem=max(tem,read(C,c-p)); 57 add(C,p,b); 58 } 59 else 60 { 61 if(p>d) 62 continue; 63 tem=read(C,c); 64 tem=max(tem,read(D,d-p)); 65 add(D,p,b); 66 } 67 if(tem) 68 an=max(an,tem+b); 69 } 70 printf("%d ",an); 71 return 0; 72 }
第一次写BIT，感觉比起线段树代码量的确少的多，但现在比较复杂的操作还不会用bit写……