Fence

Fence

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 4970		Accepted: 1565

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 

思路：這是一道dp題，首先寫出方程式：dp[i][j]=max(dp[i-1][j],dp[i][j-1],dp[i-1][k]+p[i]*(j-k))233(表示前i個人塗前j塊木板所能達到的最大收益)其中s[i]-l[i]<=k<s[i]，把式子展開後會發現dp[i][j]=max(dp[i-1][j],dp[i][j-1],dp[i-1][k]-p[i]*k+p[i]*j)因爲p[i]*j爲定值。所以只需要用一個單調隊列來維護dp[i-1][k]-p[i]*k的最小值即可把O(n^3)優化到O(N^2)~\(≧▽≦)/~

#include<iostream>

#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<queue>
#include<deque>
using namespace std;
const int N=16000+10;
const int K=100+10;
const int inf=0x3f3f3f3f;
void getint(int&num){
    char c;int flag=1;num=0;
    while((c=getchar())<'0'||c>'9')if(c=='-')flag=-1;
    while(c>='0'&&c<='9'){num=num*10+c-48;c=getchar();}
    num*=flag;
}
struct worker{
	int l,p,s;
	bool operator < (const worker &x)const{
		return s<x.s;
	}
}arr[N];
struct node{
	int num,pos;
	node(){}
	node(int a,int b){
		num=a,pos=b;
	}
};
deque<node> q;
int n,k,dp[K][N];
int main(){
	while(~scanf("%d %d",&n,&k)){
		for(int i=1;i<=k;i++)
			getint(arr[i].l),getint(arr[i].p),getint(arr[i].s);
		sort(arr+1,arr+k+1);
		memset(dp,0,sizeof(dp));
		for(int i=1;i<=k;i++){
			while(!q.empty())	q.pop_front();
			int tmp=max(arr[i].s-arr[i].l,0);
			q.push_back(node(dp[i-1][tmp]-arr[i].p*tmp,tmp));
			for(int j=1;j<=n;j++){
				dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
				if(j>=arr[i].s+arr[i].l)	continue ;
				while(!q.empty()&&q.front().pos+arr[i].l<j)
					q.pop_front();
				if(j<arr[i].s){
					int now=dp[i-1][j]-arr[i].p*j;
					while(!q.empty()&&q.back().num<now)
						q.pop_back();
					q.push_back(node(now,j));
				}
				else dp[i][j]=max(dp[i][j],q.front().num+arr[i].p*j);
			}
		}
		printf("%d\n",dp[k][n]);
	}
}