| # | 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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