题目:http://www.lydsy.com/JudgeOnline/problem.php?id=1563
打了个表发现决策具有单调性,然后就二分+队列维护即可,O(n log n)。
代码:
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std ;
#define sum( l , r ) ld( pre[ r ] - pre[ l - 1 ] )
#define REP( i , l , r ) for ( int i = l ; i <= r ; ++ i )
#define rep( i , x ) for ( int i = 0 ; i ++ < x ; )
#define Rep( i , x ) for ( int i = 0 ; i < x ; ++ i )
typedef long long ll ;
typedef long double ld ;
const int maxn = 101000 , inf = 0x7fffffff ;
const ld INF = ld( inf ) * ld( inf ) ;
const int maxl = 50 ;
const ld MAX = ld( 1000000000000000000 ) ;
int T , N , P , L , pre[ maxn ] , len[ maxn ] ;
char s[ maxl ] ;
ld power( ld x , int cnt ) {
ld y = 1 ;
for ( ; cnt ; cnt >>= 1 ) {
if ( cnt & 1 ) y *= x ;
x *= x ;
}
return y ;
}
ld dp[ maxn ] ;
ld cost( int l , int r ) {
return dp[ l - 1 ] + power( ld( abs( sum( l , r ) + ld( r - l ) - L ) ) , P ) ;
}
int q[ maxn ][ 3 ] , head = 1 , tail = 1 ;
int main( ) {
scanf( "%d" , &T ) ;
while ( T -- ) {
scanf( "%d%d%d" , &N , &L , &P ) ;
pre[ 0 ] = dp[ 0 ] = 0 ;
rep( i , N ) {
scanf( "%s" , s ) ;
pre[ i ] = pre[ i - 1 ] + ( len[ i ] = strlen( s ) ) ;
}
memset( q , 0 , sizeof( q ) ) ;
head = tail = 1 ;
q[ 1 ][ 0 ] = 1 , q[ 1 ][ 1 ] = 1 , q[ 1 ][ 2 ] = N ;
rep( i , N ) {
for ( ; q[ head ][ 2 ] < i ; ++ head ) ;
dp[ i ] = cost( q[ head ][ 0 ] , i ) ;
for ( ; head <= tail ; ) {
if ( cost( q[ tail ][ 0 ] , q[ tail ][ 1 ] ) >= cost( i + 1 , q[ tail ][ 1 ] ) ) -- tail ;
else if ( cost( q[ tail ][ 0 ] , q[ tail ][ 2 ] ) < cost( i + 1 , q[ tail ][ 2 ] ) ) break ;
else {
int l = q[ tail ][ 1 ] , r = q[ tail ][ 2 ] , mid ;
while ( r - l > 1 ) {
mid = ( l + r ) >> 1 ;
if ( cost( q[ tail ][ 0 ] , mid ) < cost( i + 1 , mid ) ) l = mid ; else r = mid ;
}
q[ tail ][ 2 ] = l ; break ;
}
}
if ( q[ tail ][ 2 ] < N ) {
q[ ++ tail ][ 0 ] = i + 1 ;
q[ tail ][ 1 ] = q[ tail - 1 ][ 2 ] + 1 , q[ tail ][ 2 ] = N ;
}
}
if ( dp[ N ] > MAX ) printf( "Too hard to arrange\n" ) ; else printf( "%lld\n" , ll( dp[ N ] ) ) ;
printf( "--------------------\n" ) ;
}
return 0 ;
}