제출 #465802

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
465802		MohamedAliSaidane	Handcrafted Gift (IOI20_gift)	C++14	45 / 100	1587 ms	18844 KiB

gift

#include <bits/stdc++.h>
#include "gift.h"
using namespace std;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef long long ll;
typedef pair<ll,ll> pll;
typedef tuple<int,int,int> ti;
typedef unsigned long long ull;
typedef long double ld;
typedef vector<ll> vll;
typedef pair<ld,ld> pld;
#define pb push_back
#define popb pop_back()
#define pf push_front
#define popf pop_front
#define ff first
#define ss second
#define MOD (int)(1e8)
#define INF (ll) (1e18)
#define all(v) (v).begin(),(v).end()
#define LSOne(S) ((S) & -(S))
ll gcd(ll a , ll b) {return b ? gcd(b , a % b) : a ;}
ll lcm(ll a , ll b) {return (a * b) / gcd(a , b);}
ld pointdist(ld x, ld y, ld point) { return ((x-point)*(y-point))/abs(x-y); }
ld dist(ld x, ld y, ld a, ld b){ return sqrt((x-a)*(x-a) + (y-b)*(y-b)); }
const int nx[4] = {0, 0, 1, -1}, ny[4] = {1, -1, 0, 0}; //East, West, South, North+
////////////******SOLUTION******\\\\\\\\\\\
 
int sz;
int req;
vi A, B, X;
string ans = "";
bool found = false;
void f(string str)
{
    if(found)
        return ;
    if(str.length() == sz)
    {
        int pref[sz+1];
        pref[0] = 0;
        for(int i = 0; i < sz; i ++)
        {
            pref[i+1] = pref[i] + (str[i] == 'R');
        }
        for(int j = 0; j < req; j ++)
        {
            int rs = pref[B[j]+1] - pref[A[j]];
            int len = B[j] + 1 - A[j];
            if(X[j] == 1)
            {
                if(rs == len || rs == 0)
                    continue;
                else
                    return ;
            }
            else
            {
                if(rs < len && rs > 0)
                    continue;
                else
                    return ;
            }
        }
        ans = str;
        found = true;
        return ;
    }
    f(str+'R');
    f(str+'B');
}
int construct(int n, int r, vi a, vi b, vi x)
{
    sz = n;
    req = r;
    A = a;
    B = b;
    X = x;
    string s(n, 'R');
    bool all_one = true;
    bool all_two = true;
    for(int i = 0; i <r; i ++)
    {
        if(x[i] != 1)
            all_one = false;
        if(x[i] != 2)
            all_two = false;
    }
    if(all_one)
    {
        craft(s);
        return 1;
    }
    else if(all_two)
    {
        for(int i = 0; i <r; i ++)
        {
            if(a[i] == b[i] || n == 1)
                return 0;
        }
        for(int j = 1; j < n; j += 2)
            s[j] = 'B';
        craft(s);
        return 1;
    }
    else
    {
        // BF
        f("");
        if(found)
        {
            craft(ans);
            return 1;
        }
        else
            return 0;
 
    }
    return 0;
 
}
 
 
/*
Identify problem diagram: Brute force, Greedy, Dynamic Programming, Divide and Conquer
Reformulate the problem into something more theoretical
!!!!! IMPLICIT GRAPH ??????
!!!!! PAY ATTENTION TO THE CONSTRAINTS: DP nD ? BF ? BITMASKING ?
!!!!! SOLVE THE SUBTASKS FIRST: IT'S TOTALLY OK NOT TO SOLVE THE PROBLEM ENTIRELY
Search for multiple approaches: select the seemingly optimal one
Remember that you're the king of CP
Change your approach
Imagine Corner cases before submitting
Don't spend too much time on the problem: move on !
*/

컴파일 시 표준 에러 (stderr) 메시지

gift.cpp:28:1: warning: multi-line comment [-Wcomment]
   28 | ////////////******SOLUTION******\\\\\\\\\\\
      | ^
gift.cpp: In function 'void f(std::string)':
gift.cpp:39:21: warning: comparison of integer expressions of different signedness: 'std::__cxx11::basic_string<char>::size_type' {aka 'long unsigned int'} and 'int' [-Wsign-compare]
   39 |     if(str.length() == sz)
      |        ~~~~~~~~~~~~~^~~~~

• Subtask 1: 10 / 10
• Subtask 2: 15 / 15
• Subtask 3: 20 / 20
• Subtask 4: 0 / 25
• Subtask 5: 0 / 30