Submission #160393

#	Time	Username	Problem	Language	Result	Execution time	Memory
160393		godwind	Dango Maker (JOI18_dango_maker)	C++14	13 / 100	3 ms	504 KiB

dango_maker

#pragma GCC optimize("Ofast")
#pragma GCC optimize("no-stack-protector")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("fast-math")
#pragma GCC target("sse,sse2,sse3,ssse3,popcnt,abm,mmx,tune=native")
#include <iostream>
#include <vector>
#include <algorithm>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <stdio.h>
#include <cstdio>
#include <math.h>
#include <cmath>
#include <string>
#include <cstring>
#include <queue>
#include <deque>
#include <random>
#include <iomanip>
#include <bitset>
 
 
using namespace std;
 
template<typename T> void uin(T &a, T b) {
    if (b < a) {
        a = b;
    }
}
 
template<typename T> void uax(T &a, T b) {
    if (b > a) {
        a = b;
    }
}
 
#define int long long
#define left left228
#define right right228
#define prev prev228
#define list list228
#define mp make_pair
#define all(v) v.begin(), v.end()
#define forn(i, n) for (int i = 0; i < (int)n; ++i)
#define firn(i, n) for (int i = 1; i < (int)n; ++i)
#define x first
#define y second

const int N = 3007;

char a[N][N];
bool used[N][N];
vector< pair<int, int> > cells[2 * N];

signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
            cin >> a[i][j];
            cells[(n - i + 1) + j].push_back({i, j});
        }
    }
    int res = 0;
    for (int s = 2; s <= n + m; ++s) {
        for (pair<int, int> p : cells[s]) {
            int i = p.first, j = p.second;
            if (j + 2 <= m && !used[i][j] && !used[i][j + 1] && !used[i][j + 2] && a[i][j] == 'R' && a[i][j + 1] == 'G' && a[i][j + 2] == 'W') {
                used[i][j] = used[i][j + 1] = used[i][j + 2] = 1;
                ++res;
            }
            if (i + 2 <= n && !used[i][j] && !used[i + 1][j] && !used[i + 2][j] && a[i][j] == 'R' && a[i + 1][j] == 'G' && a[i + 2][j] == 'W') {
                used[i][j] = used[i + 1][j] = used[i + 2][j] = 1;
                ++res;
            }
        }
    }
    cout << res << '\n';
    return 0;
}

• Subtask 1: 13 / 13
• Subtask 2: 0 / 20
• Subtask 3: 0 / 67