Codeforces 799C. Fountains

C. Fountains
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
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.

Examples
input
3 7 6
10 8 C
4 3 C
5 6 D
output
9
input
2 4 5
2 5 C
2 1 D
output
0
input
3 10 10
5 5 C
5 5 C
10 11 D
output
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.

考虑建了一个C一个D的情况 答案就是C,D中取满足cost的最大值即可
CC的情况 ， 按cost升序排序，维护sc[i]=cost小于等于i的情况下,收益最大的值
枚举每对{bi,pi}并依次插入sc中即可 CC的最大值就算sc[min(c-i,i)]+当前收益
DD同理

#include<stdio.h>
#include <iostream>
#include<stdlib.h>
#include<algorithm>
#include<vector>
#include<deque>
#include<map>
#include<set>
#include<queue>
#include<math.h>
#include<string.h>
#include<string>

using namespace std;
#define ll long long
#define pii pair<int,int>

const int inf = 1e9 + 7;

const int N = 1e5+5;

vector<pii>cf,df;
int sc[N];
int sd[N];

int slove(int n,int c,int d){
    sort(cf.begin(),cf.end());
    sort(df.begin(),df.end());
    int cc=-inf;
    int lastIdx=0;
    for(int i=0;i<cf.size();++i){
        int cost=cf[i].first;
        int score=cf[i].second;
        for(int j=lastIdx+1;j<=cost;++j){
            sc[j]=max(sc[j],sc[j-1]);
        }
        int x=min(c-cost,cost);
        if(x>=0){
            cc=max(cc,sc[x]+score);
        }
        sc[cost]=max(sc[cost],score);
        lastIdx=cost;
    }
    for(int i=lastIdx+1;i<=c;++i){
        sc[i]=max(sc[i],sc[i-1]);
    }
    int dd=-inf;
    lastIdx=0;
    for(int i=0;i<df.size();++i){
        int cost=df[i].first;
        int score=df[i].second;
        for(int j=lastIdx+1;j<=cost;++j){
            sd[j]=max(sd[j],sd[j-1]);
        }
        int x=min(d-cost,cost);
        if(x>=0){
            dd=max(dd,sd[x]+score);
        }
        sd[cost]=max(sd[cost],score);
        lastIdx=cost;
    }
    for(int i=lastIdx+1;i<=d;++i){
        sd[i]=max(sd[i],sd[i-1]);
    }
    int cd=sc[c]+sd[d];
    return max(cc,max(dd,max(0,cd)));
}

int main()
{
    //freopen("/home/lu/Documents/r.txt","r",stdin);
    //freopen("/home/lu/Documents/w.txt","w",stdout);
    int n,c,d;
    while(~scanf("%d%d%d",&n,&c,&d)){
        fill(sc,sc+c+1,-inf);
        fill(sd,sd+d+1,-inf);
        cf.clear();
        df.clear();
        int score,cost;
        char ch;
        for(int i=0;i<n;++i){
            scanf("%d%d%*c%c",&score,&cost,&ch);
            if(ch=='C'){
                cf.push_back({cost,score});
            }
            else{
                df.push_back({cost,score});
            }
        }
        printf("%d\n",slove(n,c,d));
    }
    return 0;
}