Submission #47574

#	Time	Username	Problem	Language	Result	Execution time	Memory
47574		JohnTitor	Schools (IZhO13_school)	C++11	100 / 100	229 ms	13612 KiB

school

#include <bits/stdc++.h>
using namespace std;
#define FOR(i, j, k) for(int i=(j); i<=(k); i++)
#define FFOR(i, j, k) for(int i=(j); i<(k); i++)
#define DFOR(i, j, k) for(int i=(j); i>=(k); i--)
#define bug(x) cerr<<#x<<" = "<<(x)<<'\n'
#define pb push_back
#define mp make_pair
#define setbit(s, i) (s|=(1LL<<(i)))
#define bit(s, i) (((s)>>(i))&1LL)
#define mask(i) ((1LL<<(i)))
#define builtin_popcount __builtin_popcountll
typedef long long ll;
typedef long double ld;
template <typename T> inline void read(T &x){
    char c;
    bool nega=0;
    while((!isdigit(c=getchar()))&&(c!='-'));
    if(c=='-'){
        nega=1;
        c=getchar();
    }
    x=c-48;
    while(isdigit(c=getchar())) x=x*10+c-48;
    if(nega) x=-x;
}
template <typename T> inline void writep(T x){
    if(x>9) writep(x/10);
    putchar(x%10+48);
}
template <typename T> inline void write(T x){
    if(x<0){
        putchar('-');
        x=-x;
    }
    writep(x);
}
template <typename T> inline void writeln(T x){
    write(x);
    putchar('\n');
}
#define taskname "C"
int n, m, s;
class city{
public:
    ll a, b;
} c[300001];
ll f[300001];
ll g[300001];
multiset <ll> se;
ll sum;
int main(){
    #ifdef Kanikou
        if(fopen(taskname".inp", "r"))
            freopen(taskname".inp", "r", stdin);
    #endif // Kanikou
    read(n);
    read(m);
    read(s);
    FOR(i, 1, n){
        read(c[i].a);
        read(c[i].b);
    }
    sort(c+1, c+n+1, [](city A, city B){
        return A.a-A.b>B.a-B.b;
    });
    FOR(i, 1, m){
        se.insert(c[i].a);
        sum+=c[i].a;
    }
    f[m]=sum;
    FOR(i, m+1, n){
        se.insert(c[i].a);
        sum+=c[i].a;
        sum-=*se.begin();
        se.erase(se.begin());
        f[i]=sum;
    }
    se.clear();
    sum=0;
    FOR(i, 1, s){
        se.insert(c[n-i+1].b);
        sum+=c[n-i+1].b;
    }
    g[n-s+1]=sum;
    DFOR(i, n-s, 1){
        se.insert(c[i].b);
        sum+=c[i].b;
        sum-=*se.begin();
        se.erase(se.begin());
        g[i]=sum;
    }
    ll ans=0;
    FOR(i, m, n-s) ans=max(ans, f[i]+g[i+1]);
    writeln(ans);
}

• Subtask 1: 100 / 100