Submission #107071

#	Time	Username	Problem	Language	Result	Execution time	Memory
107071		SOIVIEONE	UFO (IZhO14_ufo)	C++14	45 / 100	617 ms	47992 KiB

ufo

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>

#define INF 1000000021
#define MOD 1000000007
#define pb push_back
#define sqr(a) (a)*(a)
#define M(a, b) make_pair(a,b)
#define F first
#define S second
#define all(x) (x.begin(), x.end())
#define deb(x) cerr << #x << " = " << x << '\n'
#define N 222222

using namespace std;
using namespace __gnu_pbds;

typedef long double ld;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> indexed_set;

const ld pi = 2 * acos(0.0);
template<class T> bool umin(T& a, T b){if(a>b){a=b;return 1;}return 0;}
template<class T> bool umax(T& a, T b){if(a<b){a=b;return 1;}return 0;}
template<class T, class TT> bool pal(T a, TT n){int k=0;for(int i=0;i<=n/2;i++){if(a[i]!=a[n-i-1]){k=1;break;}}return k?0:1;}

//int month[] = {0,31,28,31,30,31,30,31,31,30,31,30,31};

int main()
{
	int n, m, r, k, p;
	cin >> n >> m >> r >> k >> p;
	ll a[n+2][m+2];
	ll dp[n + 2][m + 2];
	memset(a, 0, sizeof a);
	memset(dp, 0, sizeof dp);	
	for(int i = 1; i <= n; i ++)
	for(int j = 1; j <= m; j ++)
		cin >> a[i][j];
	while(k --)
	{
		char t;
		ll x, y;
		cin >> t >> x >> y;
		if(t == 'W')
		{
			for(int i = 1; i <= r; i ++)
			{
				a[x][i] -= y;
				if(a[x][i] < 0)
					a[x][i] = 0;
			}
		}
		else
		if(t == 'E')
		{
			for(int i = m; i >= m - r + 1; i --)
			{
				a[x][i] -= y;
				if(a[x][i] < 0)
					a[x][i] = 0;
			}
		}
		else
		if(t == 'S')
		{
			for(int i = n; i >= n - r + 1; i --)
			{
				a[i][x] -= y;
				if(a[i][x] < 0)
					a[i][x] = 0;
			}
		}
		else
		{
			for(int i = 1; i <= r; i ++)
			{
				a[i][x] -= y;
				if(a[i][x] < 0)
					a[i][x] = 0;
			}
		}
	}
	for(int i = 1; i <= n; i ++)
		for(int j = 1; j <= m; j ++)
			dp[i][j] = a[i][j] + dp[i - 1][j] + dp[i][j - 1] - dp[i - 1][j - 1];
	ll ans = 0;
	for(int i = 1; i <= n; i ++)
		for(int j = 1; j <= m; j ++)
		{
			if(i + p - 1 > n || j + p - 1 > m)
				continue;
			ll o = dp[i + p - 1][j + p - 1] - dp[i - 1][j + p - 1] - dp[i + p - 1][j - 1] + dp[i - 1][j - 1];
			umax(ans, o);
		}
	cout << ans;


	





	getchar();
	getchar();
	return 0;
	//ios::sync_with_stdio(false);
	//cin.tie(0);
}

• Subtask 1: 45 / 100