[poj1821] Fence DP 단조로운 대기열 최적화

Fence

Description
A team of k (1 <= K <= 100) workers should paint a fence which contains N (1 <= N <= 16 000) planks numbered from 1 to N from left to right. Each worker i (1 <= i <= K) should sit in front of the plank Si and he may paint only a compact interval (this means that the planks from the interval should be consecutive). This interval should contain the Si plank. Also a worker should not paint more than Li planks and for each painted plank he should receive Pi $ (1 <= Pi <= 10 000). A plank should be painted by no more than one worker. All the numbers Si should be distinct.
Being the team’s leader you want to determine for each worker the interval that he should paint, knowing that the total income should be maximal. The total income represents the sum of the workers personal income.
Write a program that determines the total maximal income obtained by the K workers. Input
The input contains: Input
N K L1 P1 S1 L2 P2 S2 … LK PK SK
Semnification
N -the number of the planks; K ? the number of the workers Li -the maximal number of planks that can be painted by worker i Pi -the sum received by worker i for a painted plank Si -the plank in front of which sits the worker i Output
The output contains a single integer, the total maximal income. Sample Input
8 4 3 2 2 3 2 3 3 3 5 1 1 7 Sample Output
17 Hint
Explanation of the sample:
the worker 1 paints the interval [1, 2];
the worker 2 paints the interval [3, 4];
the worker 3 paints the interval [5, 7];
the worker 4 does not paint any plank Source
Romania OI 2002
f[i][j]는 전 i개인을 나타내고 j의 최대치 f[i][j]=max(f[i][j-1], f[i-1][j])f[i][j]=max(f[i][j], f[i-1][k]+(j-k)*p[i])를 칠한다.f[i-1][k]+p[i]*(j-k)=(f[i-1][k]-p[i]*k)+p[i]*j 그러면 줄어드는 단조로운 대기열 유지보수 k값을 구성합니다.

#include
#include
#include
#include
using namespace std;
const int N = 16000 + 5;
int f[105][N];
struct nd{
    int l,p,s;
}pp[105];
bool cmp( nd a, nd b ){ return a.s < b.s; }
int n,k;
int q[N];
int main(){
    scanf("%d%d", &n, &k);
    for( int i = 1; i <= k; i++ ) scanf("%d%d%d", &pp[i].l, &pp[i].p, &pp[i].s );
    std::sort( pp+1, pp+k+1, cmp );
    for( int i = 1; i <= k; i++ ){
        int head = 0,tail = 0;
        q[0] = max( pp[i].s - pp[i].l, 0 );
        for( int j = 1; j <= n; j++ ){
            f[i][j] = max( f[i-1][j], f[i][j-1] );
            if( j >= pp[i].l + pp[i].s ) continue;
            while( head <= tail && q[head] + pp[i].l < j ) head++;
            if( j < pp[i].s ){
                while( head <= tail && f[i-1][q[tail]] - q[tail]*pp[i].p < f[i-1][j] - j*pp[i].p) tail--;
                q[++tail] = j;
                continue;
            }
            f[i][j] = max( f[i][j], f[i-1][q[head]] + pp[i].p*(j-q[head]));

        }
    }
    printf("%d", f[k][n]);
    return 0;
}