「雜項」Nov. 28st, 2020 - Exam. REC
阿新 • • 發佈：2020-11-28
Prob. 1
暴力為 \(\Theta(NK)\)。
正解（也許）：
把每一個全為正整數的子段找出來。
然後判斷一下中間連線的情況即可。
但是這樣決策情況太多了。
我們需要考慮貪心。
把所有整數段的個數記為 \(totP\)，每個子段的區間記為 \([posL_{i},posR_{i}]\)，區間和記為 \(sumP_{i}\)
把其他的負數段個數記為 \(totN\)，區間和記為 \(sumN_{i}\)。
當 \(totP\le k\) 答案顯然。
我們需要考慮的是 \(totP>k\) 的情況。
我們把整數段、負數段縮成點。
然後問題還是最多選 \(k\)
不過我們的序列有個性質：相鄰數的正負性不同。（gu）
好了放棄以上想法。
模擬 \(k\) 輪找全域性最大子段和，找到一次把子段乘上 \(-1\)。
#include <cstdio> typedef long long LL; const int MAXN = 1e5 + 5; int rint () { int x = 0, f = 1; char c = getchar (); for ( ; c < '0' || c > '9'; c = getchar () ) f = c == '-' ? -1 : f; for ( ; c >= '0' && c <= '9'; c = getchar () ) x = ( x << 3 ) + ( x << 1 ) + ( c & 15 ); return x * f; } template<typename _T> void wint ( _T x ) { if ( x < 0 ) putchar ( '-' ), x = ~ x + 1; if ( x > 9 ) wint ( x / 10 ); putchar ( x % 10 ^ '0' ); } template<typename _T> _T MAX ( const _T x, const _T y ) { return x < y ? y : x; } template<typename _T> void swapp ( _T &x, _T &y ) { _T w = x; x = y; y = w; } struct nodeS { LL val, dat, p, s; int l, r, pl, pr, sl, sr; nodeS ( LL V = 0, LL D = 0, LL P = 0, LL S = 0, int L = 0, int R = 0, int Pl = 0, int Pr = 0, int Sl = 0, int Sr = 0 ) { val = V, dat = D, p = P, s = S, l = L, r = R, pl = Pl, pr = Pr, sl = Sl, sr = Sr; } } nodes[MAXN * 4][2]; int n, k, a[MAXN]; bool tag[MAXN * 4]; nodeS Merge ( const nodeS lch, const nodeS rch ) { nodeS ret; ret.val = lch.val + rch.val; ret.p = MAX ( lch.p, lch.val + rch.p ); if ( ret.p == lch.p ) ret.pl = lch.pl, ret.pr = lch.pr; else ret.pl = lch.pl, ret.pr = rch.pr; ret.s = MAX ( rch.s, rch.val + lch.s ); if ( ret.s == rch.s ) ret.sl = rch.sl, ret.sr = rch.sr; else ret.sl = lch.sl, ret.sr = rch.sr; ret.dat = MAX ( lch.s + rch.p, MAX ( lch.dat, rch.dat ) ); if ( ret.dat == lch.dat ) ret.l = lch.l, ret.r = lch.r; else if ( ret.dat == rch.dat ) ret.l = rch.l, ret.r = rch.r; else ret.l = lch.sl, ret.r = rch.pr; return ret; } void Upt ( const int x ) { nodes[x][0] = Merge ( nodes[x << 1][0], nodes[x << 1 | 1][0] ); nodes[x][1] = Merge ( nodes[x << 1][1], nodes[x << 1 | 1][1] ); } void Spr ( const int x ) { if ( ! tag[x] ) return; swapp ( nodes[x << 1][0], nodes[x << 1][1] ); swapp ( nodes[x << 1 | 1][0], nodes[x << 1 | 1][1] ); tag[x << 1] ^= 1, tag[x << 1 | 1] ^= 1, tag[x] = 0; } void Build ( const int x, const int l, const int r ) { if ( l == r ) { nodes[x][0] = nodeS ( a[l], a[l], a[l], a[l], l, l, l, l, l, l ); nodes[x][1] = nodeS ( -a[l], -a[l], -a[l], -a[l], l, l, l, l, l, l ); return; } int mid = ( l + r ) >> 1; Build ( x << 1, l, mid ); Build ( x << 1 | 1, mid + 1, r ); Upt ( x ); } void Modify ( const int x, const int l, const int r, const int segL, const int segR ) { if ( l > segR || r < segL ) return; if ( l >= segL && r <= segR ) { swapp ( nodes[x][0], nodes[x][1] ); tag[x] ^= 1; return; } int mid = ( l + r ) >> 1; Spr ( x ); Modify ( x << 1, l, mid, segL, segR ); Modify ( x << 1 | 1, mid + 1, r, segL, segR ); Upt ( x ); } int main () { n = rint (), k = rint (); for ( int i = 1; i <= n; ++ i ) a[i] = rint (); Build ( 1, 1, n ); LL ans = 0; while ( k -- > 0 ) { nodeS ret = nodes[1][0]; if ( ret.dat < 0 ) break; Modify ( 1, 1, n, ret.l, ret.r ); ans += ret.dat; } wint ( ans ), putchar ( '\n' ); return 0; }
Prob. 2
設 \(f_{i,0/1}\) 表示把 \(a_{i}\) 往頭/尾放可以得到的最多的上升子序列。
\[f_{i,0}=\begin{cases}\max\{f_{j,0},f_{j,1}\}+1,a_{i}<a_{j} \\\max\{f_{j,0},f_{j,1}\},a_{i}\ge a_{j}\end{cases} \\f_{i,1}=\begin{cases}\max\{f_{j,0},f_{j,1}\},a_{i}<a_{j} \\\max\{f_{j,0},f_{j,1}\}+1,a_{i}\ge a_{j}\end{cases} \]不行。
考慮普通的 LIS 怎麼做。
\[f_{i}=\max\{f_{j}\}+1,a_{i}>a_{j} \]\[a_{i}<a_{j},i<j \]選擇往前放的元素，放得越晚越靠前。
選擇往後放的元素，放得越晚越靠後。
那麼需要做的是把相對較大的元素往後，相對較小的元素往前。
連邊，把李三花後的 \(a_{i}\) 連向 \(\text{trans}(i,0),\text{trans}(i,1),\text{trans}(x,0/1)=n-x+1/x+n\)。
#include <cstdio>
#include <algorithm>
using namespace std;
const int MAXN = 2e5 + 5;
int rint () {
int x = 0, f = 1; char c = getchar ();
for ( ; c < '0' || c > '9'; c = getchar () ) f = c == '-' ? -1 : f;
for ( ; c >= '0' && c <= '9'; c = getchar () ) x = ( x << 3 ) + ( x << 1 ) + ( c & 15 );
return x * f;
}
template<typename _T>
void wint ( _T x ) {
if ( x < 0 ) putchar ( '-' ), x = ~ x + 1;
if ( x > 9 ) wint ( x / 10 );
putchar ( x % 10 ^ '0' );
}
template<typename _T> _T MAX ( const _T x, const _T y ) { return x < y ? y : x; }
struct Value {
int val, pos;
Value ( int V = 0, int P = 0 ) { val = V, pos = P; }
bool operator < ( const Value &another ) { return val < another.val; }
} vals[MAXN];
struct GraphSet {
int to, nx;
GraphSet ( int T = 0, int N = 0 ) { to = T, nx = N; }
} as[MAXN * 2];
int n, cnt, len, degin[MAXN], a[MAXN], b[MAXN], buc[MAXN], sywf[MAXN];
void makeEdge ( const int u, const int v ) { as[++ cnt] = GraphSet ( v, degin[u] ), degin[u] = cnt; }
int Trans ( const int x, const int y ) { return ! y ? n - x + 1 : x + n; }
void ADD ( int p, const int x ) { for ( ; p <= ( n << 1 ); p += p & -p ) sywf[p] = MAX ( sywf[p], x ); }
int ASK ( int p ) { int res = 0; for ( ; p; p -= p & -p ) res = MAX ( res, sywf[p] ); return res; }
int CMP ( const int x, const int y ) { return x > y; }
int main () {
// freopen ( "dequexlis.in", "r", stdin );
// freopen ( "dequexlis.out", "w", stdout );
n = rint ();
for ( int i = 1; i <= n; ++ i ) a[i] = b[i] = rint ();
sort ( b + 1, b + 1 + n );
len = unique ( b + 1, b + 1 + n ) - b - 1;
for ( int i = 1; i <= n; ++ i ) a[i] = lower_bound ( b + 1, b + 1 + len, a[i] ) - b;
for ( int i = 1; i <= n; ++ i ) vals[i] = Value ( a[i], i );
for ( int i = 1; i <= n; ++ i ) {
makeEdge ( a[i], Trans ( i, 0 ) );
makeEdge ( a[i], Trans ( i, 1 ) );
}
int BUC = 0;
for ( int x_x = 1; x_x <= n; ++ x_x ) {
int u = x_x;
BUC = 0;
for ( int i = degin[u]; i; i = as[i].nx ) {
int v = as[i].to;
buc[++ BUC] = v;
}
sort ( buc + 1, buc + 1 + BUC, CMP );
for ( int i = 1; i <= BUC; ++ i ) ADD ( buc[i], 1 + ASK ( buc[i] - 1 ) );
}
wint ( ASK ( n << 1 ) ), putchar ( '\n' );
return 0;
}
/* Jesus bless all */
Prob. 3
放棄人生打了個 50 的記搜。
#include <cstdio>
#include <map>
#define mod ( 998244853 )
using namespace std;
template<typename _T> _T MAX ( const _T x, const _T y ) { return x < y ? y : x; }
int n, m;
namespace Course {
const int MAXN = 255;
int f[MAXN][MAXN][MAXN];
bool vis[MAXN][MAXN][MAXN];
int dfs ( const int mx, const int a, const int b ) {
if ( vis[a][b][mx] ) return f[a][b][mx];
vis[a][b][mx] = 1;
if ( a == n && b == m ) return f[a][b][mx] = mx;
if ( a < n ) f[a][b][mx] = ( f[a][b][mx] + dfs ( MAX ( mx, a - b + 1 ), a + 1, b ) ) % mod;
if ( b < m ) f[a][b][mx] = ( f[a][b][mx] + dfs ( mx, a, b + 1 ) ) % mod;
return f[a][b][mx];
}
}
int main () {
// freopen ( "maxpsum.in", "r", stdin );
// freopen ( "maxpsum.out", "w", stdout );
scanf ( "%d%d", &n, &m );
if ( n <= 250 && m <= 250 ) printf ( "%d\n", Course :: dfs ( 0, 0, 0 ) );
// else printf ( "%d\n", Might :: dfs ( 0, 0, 0 ) );
return 0;
}