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June 21, 2017 15:26
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USACO 2015 December Contest, Platinum Problem 2. High Card Low Card (Platinum)
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#include <bits/stdc++.h> | |
#define lld long long int | |
#define IZQ (nodo * 2) | |
#define DER (nodo * 2 + 1) | |
using namespace std; | |
const int MAXN = 100002; | |
const int INF = (1 << 30); | |
int n; | |
bool bucket[MAXN]; | |
int A[MAXN], B[MAXN]; | |
struct segment_tree { | |
int mini[MAXN * 5]; | |
int lazy[MAXN * 5]; | |
void propagate(int nodo, int ini, int fin) { | |
if(lazy[nodo] != INF) { | |
mini[nodo] = mini[nodo] + lazy[nodo]; | |
if(ini != fin) { | |
if(lazy[IZQ] != INF) lazy[IZQ] += lazy[nodo]; | |
else lazy[IZQ] = lazy[nodo]; | |
if(lazy[DER] != INF) lazy[DER] += lazy[nodo]; | |
else lazy[DER] = lazy[nodo]; | |
} | |
lazy[nodo] = INF; | |
} | |
} | |
int menor() { | |
propagate(1, 1, 2 * n); | |
return mini[1]; | |
} | |
void update(int nodo, int ini, int fin, int i, int j, int v) { | |
propagate(nodo, ini, fin); | |
if(ini > j || fin < i) return; | |
if(ini >= i && fin <= j) { | |
lazy[nodo] = v; | |
propagate(nodo, ini, fin); | |
} else { | |
int med = (ini + fin) / 2; | |
update(IZQ, ini, med, i, j, v); | |
update(DER, med + 1, fin, i, j, v); | |
mini[nodo] = min(mini[IZQ], mini[DER]); | |
} | |
} | |
void update(int i, int j, int v) { update(1, 1, 2 * n, i, j, v); } | |
int query(int nodo, int ini, int fin, int pos) { | |
propagate(nodo, ini, fin); | |
if(ini == fin) return mini[nodo]; | |
int med = (ini + fin) / 2; | |
if(pos <= med) return query(IZQ, ini, med, pos); | |
else return query(DER, med + 1, fin, pos); | |
} | |
int query(int pos) { | |
if(pos > 2 * n) return 0; | |
return query(1, 1, 2 * n, pos); | |
} | |
}; | |
segment_tree gananMayores, gananMenores; | |
void quita(int p) { | |
int x = gananMayores.query(p) - gananMayores.query(p + 1); | |
gananMayores.update(1, p, -x); | |
} | |
void pon(segment_tree &ST, int p, int v) { | |
ST.update(1, p, v); | |
} | |
int res; | |
void solve() { | |
for(int i = 2 * n; i > 0; i--) | |
pon(gananMayores, i, bucket[i] ? -1 : 1); | |
for(int i = n; i >= 0; i--) { | |
int enfA = i, enfB = n - enfA; | |
enfA -= max(0, -gananMayores.menor()); | |
enfB -= max(0, -gananMenores.menor()); | |
res = max(res, enfA + enfB); | |
if(i == 0) break; | |
quita(A[i]); | |
pon(gananMenores, A[i], 1); | |
pon(gananMenores, B[i], -1); | |
} | |
} | |
int main() | |
{ | |
freopen("cardgame.in", "r", stdin); | |
freopen("cardgame.out", "w", stdout); | |
ios_base::sync_with_stdio(0); cin.tie(0); | |
cin >> n; | |
for(int i = 1; i <= n; i++) { | |
cin >> A[i]; | |
bucket[A[i]] = true; | |
} | |
int aux = 0; | |
for(int i = 1; i <= 2 * n; i++) | |
if(!bucket[i]) B[++aux] = i; | |
reverse(B + 1, B + n + 1); | |
solve(); | |
cout << res << "\n"; | |
return 0; | |
} |
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