Submission #1143999

#	Time	Username	Problem	Language	Result	Execution time	Memory
1143999		monkey133	Dango Maker (JOI18_dango_maker)	C++20	100 / 100	444 ms	185576 KiB

dango_maker

#include <bits/stdc++.h>
//#include "includeall.h"
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
#define endl '\n'
#define f first
#define s second
#define pb push_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define input(x) scanf("%lld", &x);
#define input2(x, y) scanf("%lld%lld", &x, &y);
#define input3(x, y, z) scanf("%lld%lld%lld", &x, &y, &z);
#define input4(x, y, z, a) scanf("%lld%lld%lld%lld", &x, &y, &z, &a);
#define print(x, y) printf("%lld%c", x, y);
#define show(x) cerr << #x << " is " << x << endl;
#define show2(x,y) cerr << #x << " is " << x << " " << #y << " is " << y << endl;
#define show3(x,y,z) cerr << #x << " is " << x << " " << #y << " is " << y << " " << #z << " is " << z << endl;
#define all(x) x.begin(), x.end()
#define discretize(x) sort(x.begin(), x.end()); x.erase(unique(x.begin(), x.end()), x.end());
#define FOR(i, x, n) for (ll i =x; i<=n; ++i) 
#define RFOR(i, x, n) for (ll i =x; i>=n; --i) 
using namespace std;
mt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());
//using namespace __gnu_pbds;
//#define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
//#define ordered_multiset tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update>
typedef int ll;
typedef long double ld;
typedef pair<ld, ll> pd;
typedef pair<string, ll> psl;
typedef pair<ll, ll> pi;
typedef pair<ll, pi> pii;
typedef pair<pi, pi> piii;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

// compress triple into 3 in the middle 
// observe that (x, y) only effect (x - 1, y + 1) and (x + 1, y - 1) -> diagonal

ll n,m;
char a[3005][3005];
ll dp[3005][3005][3], lr[3005][3005], ud[3005][3005];

int main()
{
	cin >> n >> m;
	for (ll i=1; i<=n; ++i) for (ll j=1; j<=m; ++j) cin >> a[i][j];
	// mark valid triple
	for (ll i=1; i<=n; ++i) for (ll j=2; j<m; ++j) if (a[i][j-1]=='R' && a[i][j]=='G' && a[i][j + 1] == 'W') lr[i][j] = 1;
	for (ll j=1; j<=m; ++j) for (ll i=2; i<n; ++i) if (a[i-1][j]=='R' && a[i][j]=='G' && a[i+1][j] == 'W') ud[i][j] = 1;
	//for (ll i=1; i<=n; ++i) for (ll j=1; j<=m; ++j) show2(lr[i][j], ud[i][j]);
	for (ll i=1; i<=n; ++i) for (ll j=1; j<=m; ++j)
	{
		// choose none
		dp[i][j][0] = max(max(dp[i - 1][j + 1][0], dp[i - 1][j + 1][1]), dp[i - 1][j+1][2]);
		// choose left-right
		dp[i][j][1] = max(dp[i - 1][j + 1][0], dp[i - 1][j + 1][1]) + ll(lr[i][j]);
		// choose up-down
		dp[i][j][2] = max(dp[i - 1][j + 1][0], dp[i - 1][j + 1][2]) + ll(ud[i][j]);
		//show2(i, j);
		//show3(dp[i][j][0], dp[i][j][1], dp[i][j][2]);
	}
	ll ans = 0;
	for (ll i=1; i<=n; ++i) ans += *max_element(dp[i][1], dp[i][1] + 3);
	for (ll i=1; i<=m; ++i) ans += *max_element(dp[n][i], dp[n][i] + 3);
	ans -=  *max_element(dp[n][1], dp[n][1] + 3);
	cout << ans << endl;
	
	return 0;
}

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