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。
#include<iostream>
#include<cstdio>
#include<queue>
#define ll long long
#define RG register
using namespace std; int n, a[], b[]; priority_queue < int, vector < int > , greater < int > > q1, q2; int main ( ) {
freopen ( "buy.in", "r", stdin );
freopen ( "buy.out", "w", stdout );
scanf ( "%d", &n );
for ( RG int i = ; i <= n; i ++ )
scanf ( "%d", &a[i] );
for ( RG int i = ; i <= n; i ++ )
scanf ( "%d", &b[i] );
ll ans = ;
for ( RG int i = ; i <= n; i ++ ) {
q1.push ( a[i] );
int r1 = , r2 = ;
if ( !q1.empty ( ) ) {
int x = q1.top ( );
if ( b[i] > x ) r1 = b[i] - x;
}
if ( !q2.empty ( ) ) {
int x = q2.top ( );
if ( b[i] > x ) r2 = b[i] - x;
}
if ( r1 >= r2 && r1 ) ans += r1, q1.pop ( ), q2.push ( b[i] );
else if ( r2 > r1 && r2 ) ans += r2, q2.pop ( ), q2.push ( b[i] );
}
printf ( "%I64d", ans );
return ;
}
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。
#include<iostream>
#include<cstdio>
#define ll long long
using namespace std; int n, s, e, a[];
ll pre[], TR[]; void update ( int nd ) {
TR[nd] = max ( TR[nd << ], TR[nd << | ] );
} void build ( int nd, int l, int r ) {
if ( l == r ) {
TR[nd] = pre[l];
return ;
}
int mid = ( l + r ) >> ;
build ( nd << , l, mid );
build ( nd << | , mid + , r );
update ( nd );
} ll query ( int nd, int l, int r, int L, int R ) {
if ( l >= L && r <= R ) return TR[nd];
int mid = ( l + r ) >> ;
ll ans = -1e9;
if ( L <= mid ) ans = max ( ans, query ( nd << , l, mid, L, R ) );
if ( R > mid ) ans = max ( ans, query ( nd << | , mid + , r, L, R ) );
return ans;
} int main ( ) {
freopen ( "invest.in", "r", stdin );
freopen ( "invest.out", "w", stdout );
scanf ( "%d%d%d", &n, &s, &e );
for ( int i = ; i <= n; i ++ ) {
scanf ( "%d", &a[i] );
pre[i] = pre[i-] + a[i];
}
build ( , , n );
ll ans = ;
for ( int i = ; i <= n; i ++ ) {
if ( i + s - > n ) break;
ll x = query ( , , n, i + s - , i + e - );
ans = max ( x - pre[i-], ans );
}
printf ( "%I64d", ans );
return ;
}
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！
#include<iostream>
#include<cstdio>
#define ll long long
#define mod 19930726
using namespace std; ll max_r[];
int n;
ll k;
ll ans = , flag[]; char M[], a[]; inline ll min ( ll a, int b ) {
return a < b ? a : b;
} ll mi ( ll a, ll b ) {
ll an = ;
for ( ; b; b >>= , a = a * a % mod )
if ( b & ) an = an * a % mod;
return an;
} void manacher ( ) {
M[] = '@';
for ( int i = ; i <= n; i ++ ) {
M[ * i - ] = '#';
M[ * i] = a[i];
}
M[ * n + ] = '#'; M[ * n + ] = '$';
int center = ; ll mx = ;
int side = n * + ;
for ( int i = ; i <= n * + ; i ++ ) {
if ( mx > i ) max_r[i] = min ( mx - (ll)i, max_r[center * - i] );
else max_r[i] = ;
while ( M[max_r[i]+i] == M[i-max_r[i]] ) max_r[i] ++;
if ( mx < i + max_r[i] ) {
mx = i + max_r[i]; center = i;
}
}
} int main ( ) {
freopen ( "rehearse.in", "r", stdin );
freopen ( "rehearse.out", "w", stdout );
scanf ( "%d%I64d\n", &n, &k );
scanf ( "%s", a + );
manacher ( );
ll MA = ;
for ( int i = ; i <= n; i ++ ) {
max_r[i*] --;
flag[max_r[i*]] ++;
MA = max ( MA, max_r[i*] );
}
ll pos = MA;
while ( k > ) {
ans = ( ans * mi ( pos, min ( flag[pos], k ) ) ) % mod;
flag[pos-] += flag[pos];
k -= flag[pos];
pos = pos - ;
}
printf ( "%I64d", ans );
return ;
}