제출 #1072121

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1072121		vjudge1	Double Attendance (CCO22_day1problem3)	C++14	21 / 25	79 ms	63368 KiB

Main

#include<bits/stdc++.h>
#include<fstream>
using namespace std;
#define sz(a) (int)a.size()
#define ALL(v) v.begin(), v.end()
#define ALLR(v) v.rbegin(), v.rend()
#define ll long long
#define pb push_back
#define forr(i, a, b) for(int i = a; i < b; i++)
#define dorr(i, a, b) for(int i = a; i >= b; i--)
#define ld long double
#define vt vector
#include<fstream>
#define fi first
#define se second
#define pll pair<ll, ll>
#define pii pair<int, int>
#define mpp make_pair
const ld PI = 3.14159265359, prec = 1e-9;;
//using u128 = __uint128_t;
//const int x[4] = {1, 0, -1, 0};
//const int y[4] = {0, -1, 0, 1};
const ll mod =1e9 + 7, pr = 31;
const int mxn = 1e6 + 5, mxq = 1e5 + 5, sq = 500, mxv = 5e4 + 1;
//const int base = (1 <<18);
const ll inf = 1e9 + 5, neg = -69420, inf2 = 1e14;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
// have fun!
int n1, n2, t;
void comb(pii &a, pii b){
    if(a.fi < b.fi)a = b;
    if(a.fi == b.fi && a.se > b.se)a = b;
}
pii dp[2005][2005][2], one[2005], two[2005];
void solve(){
    cin >> n1 >> n2 >> t;
    for(int i = 1; i <= n1; i++){
        cin >> one[i].fi >> one[i].se;
    }
    sort(one + 1, one + n1 + 1);
    for(int i = 1; i <= n2; i++){
        cin >> two[i].fi >> two[i].se;
    }
    sort(two + 1, two + n2 + 1);
    for(int i = 0; i <= n1; i++){
        for(int j = 0; j <= n2; j++){
            dp[i][j][0] = dp[i][j][1] = mpp(-inf, inf);
        }
    }
    dp[0][0][0] = mpp(0, 0);
    int ans = 0;
    for(int i = 0; i <= n1; i++){
        for(int j = 0; j <= n2; j++){
            for(int k = 0; k < 2; k++){
                if(dp[i][j][k].fi == -inf)continue;
                if(k == 0){
                    if(i < n1){
                        assert(dp[i][j][k].se <= one[i + 1].fi);
                        comb(dp[i + 1][j][0], mpp(dp[i][j][k].fi + 1, one[i + 1].fi));
                    }if(j < n2){
                        comb(dp[i][j + 1][0], dp[i][j][k]);
                        if(dp[i][j][k].se + t < two[j + 1].se){
                            comb(dp[i][j + 1][1], mpp(dp[i][j][k].fi + 1, max(dp[i][j][k].se + t, two[j + 1].fi)));
                        }
                    }
                }else{
                    if(j < n2){
                        assert(dp[i][j][k].se <= two[j + 1].fi);
                        comb(dp[i][j + 1][1], mpp(dp[i][j][k].fi + 1, two[j + 1].fi));
                    }if(i < n1){
                        comb(dp[i + 1][j][1], dp[i][j][k]);
                        if(dp[i][j][k].se + t < one[i + 1].se){
                            comb(dp[i + 1][j][0], mpp(dp[i][j][k].fi + 1, max(dp[i][j][k].se + t, one[i + 1].fi)));
                        }
                    }
                }
                ans = max(ans, dp[i][j][k].fi);
            }
        }
    }
    
    cout << ans;
}
 
 
 
signed main(){
    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    //freopen("THREE.inp", "r", stdin);
    //freopen("THREE.out", "w", stdout);
    int tt; tt = 1;
    while(tt--){
        solve();
 
    }
    return(0);
}

• Subtask 1: 5 / 5
• Subtask 2: 10 / 10
• Subtask 3: 6 / 6
• Subtask 4: 0 / 4