제출 #220402

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
220402		summitwei	로봇 (IOI13_robots)	C++17	90 / 100	3056 ms	45304 KiB

robots

#include <bits/stdc++.h>
#include <robots.h>
using namespace std;
typedef vector<int> vi;
typedef vector<pair<int, int> > vpii;
typedef pair<int, int> pii;
typedef long long ll;
typedef pair<ll, ll> pll;
typedef pair<double, double> pdd;
typedef vector<ll> vll;
#define INF 0x3f3f3f3f
#define MOD 1000000007LL
#define EPSILON 0.00001
#define f first
#define s second
#define pb push_back
#define mp make_pair

#define FOR(i, a, b) for (ll i=(a); i<=(signed)(b); i++)
#define F0R(i, a) for (ll i=0; i<(signed)(a); i++)
#define RFOR(i, a, b) for (ll i=(a); i >= b; i--)

#define MN 1000005
int n, m, t;
pii coo[MN];
int vert[MN];
int horiz[MN];
map<int, int> vcmp, hcmp;
int tim1, tim2;

ll tree[2*MN]; //range modify, store global min
ll lazy[2*MN];
void upd(int node, int a, int b, int i, int j, ll val){
    if(lazy[node] != 0){
        tree[node] += lazy[node];
        if(a != b){
            lazy[node*2] += lazy[node];
            lazy[node*2+1] += lazy[node];
        }
        lazy[node] = 0;
    }
    if(a > j || b < i) return;
    if(i <= a && b <= j){
        tree[node] += val;
        if(a != b){
            lazy[node*2] += val;
            lazy[node*2+1] += val;
        }
        return;
    }
    int mid = (a+b)/2;
    upd(node*2, a, mid, i, j, val);
    upd(node*2+1, 1+mid, b, i, j, val);
    tree[node] = min(tree[node*2], tree[node*2+1]);
}
void pr(int node, int a, int b){
    cout << node << " " << a << " " << b << " " << tree[node] << " " << lazy[node] << "\n";
    if(a == b) return;
    pr(node*2, a, (a+b)/2);
    pr(node*2+1, 1+(a+b)/2, b);
}

bool chk(int x){
    //cout << "checking " << x << "\n";
    //cout << "tim2 " << tim2 << "\n";
    memset(tree, 0, sizeof tree);
    memset(lazy, 0, sizeof lazy);

    F0R(i, m){
        //cout << "upd 1 " << horiz[i] << " " << x << "\n";
        upd(1, 1, tim2+1, 1, horiz[i], x);
    }
    //pr(1, 1, tim2+1);
    int cur = t-1;
    RFOR(i, n-1, 0){
        while(cur >= 0 && coo[cur].f > vert[i]){
            //cout << "upd " << coo[cur].s << " " << coo[cur].s << " -1\n";
            upd(1, 1, tim2+1, 1, coo[cur].s, -1);
            --cur;
        }
        //pr(1, 1, tim2+1);
        if(tree[1] < 0) return false;
        //cout << "upd 1 " << tim2+1 << " " << x << "\n";
        upd(1, 1, tim2+1, 1, tim2+1, x);
    }
    while(cur >= 0){
        //cout << "upd " << coo[cur].s << " " << coo[cur].s << " -1\n";
        upd(1, 1, tim2+1, 1, coo[cur].s, -1);
        --cur;
    }
    if(tree[1] < 0) return false;
    return true;
}
int putaway(int A, int B, int T, int X[], int Y[], int W[], int S[]){
    n = A; m = B; t = T;
    F0R(i, n){
        vert[i] = X[i]-1;
        vcmp[vert[i]] = 0;
    }
    F0R(i, m){
        horiz[i] = Y[i]-1;
        hcmp[horiz[i]] = 0;
    }
    //int tim1 = 0;
    for(auto &u : vcmp) u.s = ++tim1;
    //int tim2 = 0;
    for(auto &u : hcmp) u.s = ++tim2;
    F0R(i, n)
        vert[i] = vcmp[vert[i]];
    F0R(i, m)
        horiz[i] = hcmp[horiz[i]];

    F0R(i, t){
        //cout << "doing " << i << "\n";
        if(vcmp.lower_bound(W[i]) == vcmp.end()) coo[i].f = tim1+1;
        else coo[i].f = vcmp.lower_bound(W[i])->s;
        if(hcmp.lower_bound(S[i]) == hcmp.end()) coo[i].s = tim2+1;
        else coo[i].s = hcmp.lower_bound(S[i])->s;

        if(coo[i].f == tim1+1 && coo[i].s == tim2+1) return -1;
        //cout << coo[i].f << " " << coo[i].s << "\n";
    }

    sort(vert, vert+n);
    sort(coo, coo+t);
    //F0R(i, t) cout << coo[i].f << " " << coo[i].s << "\n";

    int l = 0, r = t;
    while(l+1 < r){
        int mid = (l+r)/2;
        if(chk(mid)){
            r = mid;
        } else{
            l = mid;
        }
    }
    return r;
}

/*int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(0);

    int A, B, T;
    cin >> A >> B >> T;
    int X[A], Y[B], W[T], S[T];
    F0R(i, A) cin >> X[i];
    F0R(i, B) cin >> Y[i];
    F0R(i, T) cin >> W[i] >> S[i];
    cout << putaway(A, B, T, X, Y, W, S) << "\n";
    
    return 0;
}*/

• Subtask 1: 14 / 14
• Subtask 2: 14 / 14
• Subtask 3: 25 / 25
• Subtask 4: 37 / 37
• Subtask 5: 0 / 10