# | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
---|---|---|---|---|---|---|---|
88226 | kjain_1810 | Maja (COCI18_maja) | C++11 | 233 ms | 1088 KiB |
이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
#define pb push_back
#define f first
#define s second
#define ind(a) scanf("%d", &a)
#define inlld(a) scanf("%lld", &a)
#define ind2(a, b) scanf("%d%d", &a, &b)
#define inlld2(a, b) scanf("%lld%lld", &a, &b)
#define ind3(a, b, c) scanf("%d%d%d", &a, &b, &c)
#define inlld3(a, b, c) scanf("%lld%lld%lld", &a, &b, &c)
using namespace std;
const int N=105+5;
const int MOD=1e9+7;
typedef long long ll;
typedef long double ld;
ll n, m, stx, sty, k, dp[2][N][N], arr[N][N];
int main()
{
inlld2(n, m);
inlld2(stx, sty);
inlld(k);
for(ll a=1; a<=n; a++)
for(ll b=1; b<=m; b++)
{
inlld(arr[a][b]);
for(ll c=0; c<=n+m; c++)
dp[c][a][b]=-1e18;
}
dp[0][stx][sty]=0;
for(ll steps=1; steps<=min(k/2, n*m-1); steps++)
for(ll a=1; a<=n; a++)
for(ll b=1; b<=m; b++)
{
if(a>1)
dp[steps%2][a][b]=max(dp[steps%2][a][b], dp[(steps-1)%2][a-1][b]+arr[a][b]);
if(b>1)
dp[steps%2][a][b]=max(dp[steps%2][a][b], dp[(steps-1)%2][a][b-1]+arr[a][b]);
if(a<n)
dp[steps%2][a][b]=max(dp[steps%2][a][b], dp[(steps-1)%2][a+1][b]+arr[a][b]);
if(b<m)
dp[steps%2][a][b]=max(dp[steps%2][a][b], dp[(steps-1)%2][a][b+1]+arr[a][b]);
// printf("%lld %lld %lld %lld\n", steps, a, b, dp[steps][a][b]);
}
if(k/2<=n*m-1)
{
ll ans=0;
for(ll a=1; a<=n; a++)
for(ll b=1; b<=m; b++)
ans=max(ans, 2*dp[(k/2)%2][a][b]-arr[a][b]);
printf("%lld\n", ans);
return 0;
}
ll ans=0;
for(ll a=1; a<=n; a++)
for(ll b=1; b<=m; b++)
{
ll here=dp[(n*m-1)%2][a][b];
ll toadd=0;
if(a>1)
toadd=max(toadd, arr[a-1][b]);
if(a<n)
toadd=max(toadd, arr[a+1][b]);
if(b<m)
toadd=max(toadd, arr[a][b+1]);
if(b>1)
toadd=max(toadd, arr[a][b-1]);
ans=max(ans, 2*here+((k-2*n*m+2)*(toadd+arr[a][b]))/2-arr[a][b]);
}
printf("%lld\n", ans);
}
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