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", &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 }
原文地址：https://www.cnblogs.com/wans-caesar-02111007/p/9484047.html
时间： 2024-08-29 23:44:06