# [POJ 1821] Fence【DP+单调队列优化】

• 2018-01-24
• 0
• 0

## Problem:

 Time Limit: 1000MS Memory Limit: 30000K

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
Signification
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

## Solution:

• dp[ i ][ j ] = max{ dp[ i ][ j - 1 ], dp[ i - 1 ][ j ], dp[ i - 1 ][ k ] + p[ i ] * ( j - k ) }
• max 中的第一项表示第 j 块木板不涂；
• 第二项表示第 i 个油漆工不工作；
• 第三项表示前 i - 1 个油漆工涂了前 k 块，第 i 个涂了 j - k 块，那么
• j 的取值范围为 [ Si, Si + Li ) …… (1)
• k 的取值范围为 [ j - Li, min{ j, Si } ) …… (2)

## Code: O(KN) [7588K, 219MS]

```#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;

int N, K;
int dp[110][16010];
int q[16010], fr, re;

struct Worker{
int L, P, S;

inline bool operator < (const Worker &w2) const {return S < w2.S;}
} w[110];

int main(){
scanf("%d%d", &N, &K);
for(register int i = 1; i <= K; i++) scanf("%d%d%d", &w[i].L, &w[i].P, &w[i].S);
sort(w + 1, w + K + 1);
memset(dp, 0, sizeof(dp));
for(register int i = 1; i <= K; i++){
fr = re = 0, q[re++] = max(w[i].S - w[i].L, 0);
for(register int j = 1; j <= N; j++){
dp[i][j] = max(dp[i][j - 1], dp[i - 1][j]);  // This transformation must be done !!!!!!
if(j < w[i].S || j >= w[i].S + w[i].L) continue;
int lb = j - w[i].L, ub = min(j, w[i].S) - 1;
#define calc(x) (dp[i - 1][x] - w[i].P * (x))
// Maintain the monotone queue using a j-uncorrelated method !!!
while(q[re - 1] + 1 <= ub){
int cur = q[re - 1] + 1;
while(fr != re && calc(cur) >= calc(q[re - 1])) re--;
q[re++] = cur;
}
while(fr != re && q[fr] < lb) fr++;
if(fr != re) dp[i][j] = max(dp[i][j], calc(q[fr]) + w[i].P * j);
}
}
printf("%d\n", dp[K][N]);
return 0;
}
```

#### 评论

darkleafin.cf
(该域名已过期且被抢注。。)
darkleafin.github.io

49750

https://github.com/Darkleafin

OPEN AT 2017.12.10

Please refresh the page if the code cannot be displayed normally.

https://visualgo.net/en

- Theme by Qzhai