이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
using namespace std;
typedef long long ll;
int n, m, x1, y1, k, odkud[5][105][105];
ll dp[5][105][105], c[105][105], sol;
const ll inf = 1e18;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, 1, 0, -1};
int main ()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> n >> m >> x1 >> y1 >> k;
for(int i = 1 ; i <= n ; ++i){
for(int j = 1 ; j <= m ; ++j){
cin >> c[i][j];
}
}
for(int i = 1 ; i <= n ; ++i){
for(int j = 1 ; j <= m ; ++j){
for(int v = 0 ; v < 5 ; ++v){
dp[v][i][j] = -inf;
}
}
}
for(int i = 0 ; i < 5 ; ++i){
dp[i][x1][y1] = 0;
}
memset(odkud, -1, sizeof(odkud));
for(int i = 1 ; i <= min(k, 3 * n * m) ; ++i){
for(int j = 1 ; j <= n ; ++j){
for(int v = 1 ; v <= m ; ++v){
for(int w = 0 ; w < 4 ; ++w){
int nx = j + dx[w];
int ny = v + dy[w];
if(nx <= 0 || nx > n || ny <= 0 || ny > m)
continue;
if(dp[i % 5][nx][ny] < dp[(i - 1 + 5) % 5][j][v] + c[nx][ny]){
odkud[i % 5][nx][ny] = w;
dp[i % 5][nx][ny] = dp[(i - 1 + 5) % 5][j][v] + c[nx][ny];
}
}
}
}
for(int j = 1 ; j <= n ; ++j){
for(int v = 1 ; v <= m ; ++v){
if(odkud[i % 5][j][v] == odkud[(i - 2 + 5) % 5][j][v]){
int j1 = j - dx[odkud[i % 5][j][v]], v1 = v - dy[odkud[i % 5][j][v]];
sol = max(sol, 2 * dp[(i - 2 + 5) % 5][j][v] + ((c[j][v] + c[j1][v1]) * ((k - (i - 2) * 2) / 2)) - c[j][v]);
}
}
}
}
cout << (k <= 3 * n * m ? dp[k & 5][x1][y1] : sol) << endl;
return 0;
}
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