8.15校内测试队列manacher

Posted 2020-12-30 wans-caesar-02111007

dp？？不能确定转移状态。考虑用优先队列储存最优决策点，可是发现当前选择最优不能保证最后最优，在后面可以将之前用过的替换过来。

比如数据：

3 5

4 6

只储存a[i]来决策不能延展到后面的状态，因此每次选择过后把b[i]加入队列，下次选择最优时如果选择到了b[i]，则表示用之前选择过的来替换到当前状态。

这里我开了两个优先队列。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<queue>
 4 #define ll long long
 5 #define RG register
 6 using namespace std;
 7 
 8 int n, a[100005], b[100005];
 9 
10 priority_queue < int, vector < int > , greater < int > > q1, q2;
11 
12 int main ( ) {
13     freopen ( "buy.in", "r", stdin );
14     freopen ( "buy.out", "w", stdout );
15     scanf ( "%d", &n );
16     for ( RG int i = 1; i <= n; i ++ )
17         scanf ( "%d", &a[i] );
18     for ( RG int i = 1; i <= n; i ++ )
19         scanf ( "%d", &b[i] );
20     ll ans = 0;
21     for ( RG int i = 1; i <= n; i ++ ) {
22         q1.push ( a[i] );
23         int r1 = 0, r2 = 0;
24         if ( !q1.empty ( ) ) {
25             int x = q1.top ( );
26             if ( b[i] > x ) r1 = b[i] - x;
27         }
28         if ( !q2.empty ( ) ) {
29             int x = q2.top ( );
30             if ( b[i] > x ) r2 = b[i] - x;
31         }

32         if ( r1 >= r2 && r1 ) ans += r1, q1.pop ( ), q2.push ( b[i] );
33         else if ( r2 > r1 && r2 ) ans += r2, q2.pop ( ), q2.push ( b[i] );
34     }
35     printf ( "%I64d", ans );
36     return 0;
37 }

记录前缀和，可以发现，从某一个点为起点时，向后延展出去的长度中一定有i到i+s这一段，所以用前缀和最大值建一棵线段树，每次查找i+s-1到i+e-1段的最大值，减去i-1的前缀和比较答案即可。

 1 #include<iostream>
 2 #include<cstdio>
 3 #define ll long long 
 4 using namespace std;
 5 
 6 int n, s, e, a[100005];
 7 ll pre[100005], TR[400005];
 8 
 9 void update ( int nd ) {
10     TR[nd] = max ( TR[nd << 1], TR[nd << 1 | 1] );
11 }
12 
13 void build ( int nd, int l, int r ) {
14     if ( l == r ) {
15         TR[nd] = pre[l];
16         return ;
17     }
18     int mid = ( l + r ) >> 1;
19     build ( nd << 1, l, mid );
20     build ( nd << 1 | 1, mid + 1, r );
21     update ( nd );
22 }
23 
24 ll query ( int nd, int l, int r, int L, int R ) {
25     if ( l >= L && r <= R ) return TR[nd];
26     int mid = ( l + r ) >> 1;
27     ll ans = -1e9;
28     if ( L <= mid ) ans = max ( ans, query ( nd << 1, l, mid, L, R ) );
29     if ( R > mid ) ans = max ( ans, query ( nd << 1 | 1, mid + 1, r, L, R ) );
30     return ans;
31 }
32 
33 int main ( ) {
34     freopen ( "invest.in", "r", stdin );
35     freopen ( "invest.out", "w", stdout );
36     scanf ( "%d%d%d", &n, &s, &e );
37     for ( int i = 1; i <= n; i ++ ) {
38         scanf ( "%d", &a[i] );
39         pre[i] = pre[i-1] + a[i];
40     }
41     build ( 1, 1, n );
42     ll ans = 0;
43     for ( int i = 1; i <= n; i ++ ) {
44         if ( i + s - 1 > n ) break;
45         ll x = query ( 1, 1, n, i + s - 1, i + e - 1 );
46         ans = max ( x - pre[i-1], ans );
47     }
48     printf ( "%I64d", ans );
49     return 0;
50 }

关键时候manacher忘了怎么写！！先manacher一遍处理出以每个点为中心点的最长回文串长度，一定是奇数。开桶记录每个长度出现次数，从大到小枚举长度l，每次把l-2的次数加上l的次数，因为l的长度满足回文串l-2一定满足（同一中心点，注意k要开long long！

 1 #include<iostream>
 2 #include<cstdio>
 3 #define ll long long
 4 #define mod 19930726
 5 using namespace std;
 6 
 7 ll max_r[2000005];
 8 int n;
 9 ll k;
10 ll ans = 1, flag[1000005];
11 
12 char M[2000005], a[1000005];
13 
14 inline ll min ( ll a, int b ) {
15     return a < b ? a : b;
16 }
17 
18 ll mi ( ll a, ll b ) {
19     ll an = 1;
20     for ( ; b; b >>= 1, a = a * a % mod )
21         if ( b & 1 ) an = an * a % mod;
22     return an;
23 }
24 
25 void manacher ( ) {
26     M[0] = ‘@‘;
27     for ( int i = 1; i <= n; i ++ ) {
28         M[2 * i - 1] = ‘#‘;
29         M[2 * i] = a[i];
30     }
31     M[2 * n + 1] = ‘#‘; M[2 * n + 2] = ‘$‘;
32     int center = 0; ll mx = 0;
33     int side = n * 2 + 1;
34     for ( int i = 1; i <= n * 2 + 1; i ++ ) {
35         if ( mx > i ) max_r[i] = min ( mx - (ll)i, max_r[center * 2 - i] );
36         else max_r[i] = 1;
37         while ( M[max_r[i]+i] == M[i-max_r[i]] ) max_r[i] ++;
38         if ( mx < i + max_r[i] ) {
39             mx = i + max_r[i]; center = i;
40         }
41     }
42 }
43 
44 int main ( ) {
45     freopen ( "rehearse.in", "r", stdin );
46     freopen ( "rehearse.out", "w", stdout );
47     scanf ( "%d%I64d
", &n, &k );
48     scanf ( "%s", a + 1 );
49     manacher ( );
50     ll MA = 0;
51     for ( int i = 1; i <= n; i ++ ) {
52         max_r[i*2] --;
53         flag[max_r[i*2]] ++;
54         MA = max ( MA, max_r[i*2] );
55     }
56     ll pos = MA;
57     while ( k > 0 ) {
58         ans = ( ans * mi ( pos, min ( flag[pos], k ) ) ) % mod;
59         flag[pos-2] += flag[pos];
60         k -= flag[pos];
61         pos = pos - 2;
62     }
63     printf ( "%I64d", ans );
64     return 0;
65 }