poj 1821 Fence(单调队列+dp)_poj1821-优快云博客

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本文深入探讨了AI音视频处理领域中的视频分割与语义识别技术，介绍了如何通过计算机视觉算法实现对视频内容的智能解析与理解。包括视频分割的方法、语义识别的应用场景及其实现过程，旨在为开发者提供有效的解决方案，提升音视频处理的智能化水平。

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Language:

Fence

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 4161		Accepted: 1288

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

Sample Output

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

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define bug printf("hihi\n")

#define eps 1e-8
typedef __int64 ll;

using namespace std;

#define N 16005
int dp[105][N];

struct stud{
  int len,pos,price;
  bool operator <(const stud b) const
  {
      return pos<b.pos;
  }
}f[N];

int n,len;
int que[N];
int tail,head;

void solve()
{
    int i,j;
    memset(dp,0,sizeof(dp));
    for(i=1;i<=n;i++)
    {
        for(int j=0;j<=len;j++)
            dp[i][j]=dp[i-1][j];

        int le,pos,price;
        le=f[i].len;
        pos=f[i].pos;
        price=f[i].price;
        head=tail=0;
        for(j=0;j<=len;j++)
        {
            if(pos+le<=j) break;
            if(j<pos)
            {
                int temp=dp[i-1][j]-j*price;
                while(head<tail&&dp[i-1][que[tail-1]]-que[tail-1]*price<temp)
                    tail--;
                que[tail++]=j;
                continue;
            }
            while(head<tail&&que[head]+le<j) head++;
            dp[i][j]=max(dp[i][j],dp[i-1][que[head]]-que[head]*price+j*price);
        }
        for(j=1;j<=len;j++)
            dp[i][j]=max(dp[i][j],dp[i][j-1]);
    }
    printf("%d\n",dp[n][len]);
}

int main()
{
    int i,j;
    while(~scanf("%d%d",&len,&n))
    {
        for(i=1;i<=n;i++)
            scanf("%d%d%d",&f[i].len,&f[i].price,&f[i].pos);
        sort(f+1,f+n+1);
        solve();
    }
    return 0;
}