| # | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
|---|---|---|---|---|---|---|---|
| 69641 | 2018-08-21T10:29:01 Z | 3zp | Homecoming (BOI18_homecoming) | C++14 | 0 ms | 0 KB |
#include<bits/stdc++.h>
#include "homecoming.h"
using namespace std;
long long sa[6000009], sb[6000009];
long long su (long long l, long long r){
if(!l) return -sb[r]+sa[r];
return -sb[r] + sa[r] + sb[l-1] - sa[l-1];
}
long long M[4000009][22];
long long dp[4000009][2];
long long solve(long long N, long long K, long long *A, long long *B){
for(long long i = 0; i < 2*N; i++)
A[i+N] = A[i],
B[i+N] = B[i];
sa[0] = A[0];
sb[0] = B[0];
for(long long i = 1; i < 3*N; i++)
sa[i] = sa[i-1] + A[i],
sb[i] = sb[i-1] + B[i];
for(int i = 0 ;i < 2*N; i++)
M[i][0] = su(i,i);
for(long long i = 1; (1 << i) < 2*N; i++){
for(long long j = 0; j + (1<<i) - 1 < 2*N; j++){
long long m1 = M[j][i-1];
long long m2 = M[j+(1<<(i-1))][i-1];
M[j][i] = max(m2, m1 + su(j+(1<<i-1), j+(1<<i) - 1));
}
}
long long masof = 1e9, FI = -1;
for(long long i = N; i < 2*N; i++){
long long S = A[i] - sb[i+K-1] + sb[i-1];
long long x = i, ss = 0;
for(long long L = 21; L >= 0; L--){
if(x - (1<<L) < 0) continue;
//cout << ss <<" "<< x- (1<<L) <<" "<<L<<" "<<M[x-(1<<L)][L]<<" "<<S<<endl;
if(M[x - (1<<L)][L] + ss + S >= 0) continue;
ss += su(x - (1 << L), x -1);
x = x - (1 << L);
}
int l;
if(S >= 0) l = i;
else l = x - 1;
if(l <= i - N) l = -1;
if(l != -1 && masof > i - l) {
FI = i;
masof = i - l;
}
// cout << i <<" "<<l << " "<<S<<" "<<A[i]<<" "<<sb[i+K-1]<<" "<<sb[i-1]<<endl;
}
if(FI == -1) return 0;
FI -= N;
dp[FI][0] = -1e18;
dp[FI][1] = A[FI] - sb[K+FI-1] + sb[FI-1];
for(long long i = FI + 1; i < FI + N - K; i++){
dp[i][0] = dp[i-1][0] + dp[i-1][1];
dp[i][1] = dp[i-1][1] + A[i] - B[i + K - 1];
if(i - K >= FI) dp[i][1] =max(dp[i][1], dp[i - K][0] + dp[i - K][1] +
A[i] - sb[i+ K - 1] + sb[i-1]);
}
long long ans = max(dp[FI + N - K - 1][0], dp[FI+ N - K - 1][1]);
long long s = 0;
for(long long i = FI - 1; i > FI - N + K; i --){
s += A[i+N] - B[i+N];
ans =max(ans, s + max(dp[i + N - K - 1][0], dp[i + N - K - 1][1]));
}
return max(max((long long)0,ans), sa[N-1]-sb[N-1]);
}