Zoj 2501-A mini locomotive

Question: There is a string of numbers, and m different intervals are extracted from them. The length of each interval cannot exceed M, so that all numbers and the maximum values are obtained.

Analysis: DP, monotonous queue, maximum range field and. Because the data is positive and does not need to beMonotonous queue maintenance (otherwise use).

The maximum range field and, find the first K items and each element as the end mark; take the end position as the DP state;

Then, the Maintenance Interval Length of the monotonic queue is used, and the O (1) time is used to find the first J items that meet the minimum length and the difference is enough.

Status: set f (I, j) to the first J number, take the largest sum of the I ranges, and S (j) to the maximum single interval field and of the first J numbers;

Transfer: f (I, j) = max (f (I-1, k) + S (j) {where K ≤ j-m }.

(If the data can be negative and the equation remains unchanged, use a monotonic queue to calculate the maximum field sum of a single interval)

Note: The data is positive. You can use the simpler equation DP... )

#include <stdio.h>#include <stdlib.h>int main(){    int F[ 4 ][ 50005 ];    int T,N,M,t,i,m,l,r,s;    while ( ~scanf("%d",&T) )    for ( t = 1 ; t <= T ; ++ t ) {        scanf("%d",&N);        for ( i = 1 ; i <= N ; ++ i )            scanf("%d",&F[ 0 ][ i ]);        scanf("%d",&M);                for ( s = 0,l = i = 1 ; i <= N ; ++ i ) {            s += F[ 0 ][ i ];             if ( i-l >= M ) s -= F[ 0 ][ l ++ ];            F[ 1 ][ i ] = s;        }                for ( m = 2 ; m <= 3 ; ++ m )        for ( s = 0,i = 1 ; i <= N ; ++ i ) {            if ( s < F[ m-1 ][ i-M ] )                 s = F[ m-1 ][ i-M ];            F[ m ][ i ] = s + F[ 1 ][ i ];        }        int max = 0;        for ( i = 1 ; i <= N ; ++ i )            if ( max < F[ 3 ][ i ] )                max = F[ 3 ][ i ];                        printf("%d\n",max);        }    return 0;}

Zoj 2501-A mini locomotive