poj 3245 Sequence Partitioning（dp+二分+单调队列+sbt）

最新推荐文章于 2020-06-19 10:47:42 发布

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最新推荐文章于 2020-06-19 10:47:42 发布

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分类专栏： DP 文章标签： input output pair bi each

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该博客探讨了POJ 3245题目，涉及数列分割问题。博主提出了解决策略，即根据数列元素的关系进行动态规划合并，并利用二分查找与单调队列来满足限制条件，以找到使B元素和最大化的分段方法。同时，博主提供了问题分析和代码实现。

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Sequence Partitioning

Time Limit: 8000MS		Memory Limit: 65536K
Total Submissions: 710		Accepted: 207
Case Time Limit: 5000MS

Description

Given a sequence of N ordered pairs of positive integers (A_i, B_i), you have to partition it into several contiguous parts. Let p be the number of these parts, whose boundaries are (l₁, r₁), (l₂, r₂), ... ,(l_p, r_p), which satisfy l_i=r_{i −}₁+ 1, l_i≤ r_i, l₁= 1, r_p= n. The parts themselves also satisfy the following restrictions:

For any two pairs (A_p, B_p), (A_q, B_q), where (A_p, B_p) is belongs to the T_pth part and (A_q, B_q) the T_qth part. If T_p < T_q, then B_p> A_q.
Let M_i be the maximum A-component of elements in the ith part, say

M_i = max{A_{l_i}, A_{l_i+}_₁, ..., A_{r_i}}, 1 ≤ i ≤ p

it is provided that

where Limit is a given integer.

Let S_i be the sum of B-components of elements in the ith part. Now I want to minimize the value

max{S_i:1 ≤ i ≤ p}

Could you tell me the minimum?

Input

The input contains exactly one test case. The first line of input contains two positive integers N (N ≤ 50000), Limit (Limit ≤ 2³¹-1). Then follow N lines each contains a positive integers pair (A, B). It's always guaranteed that

max{ A ₁, A ₂, ..., A_n} ≤ Limit

Output

Output the minimum target value.

Sample Input

Sample Output

Hint

An available assignment is the first two pairs are assigned into the first part and the last two pairs are assigned into the second part. Then B ₁ > A ₃, B ₁ > A ₄, B ₂ > A ₃, B ₂ > A ₄, max{ A ₁, A ₂}+max{ A ₃, A ₄} ≤ 6, and minimum max{ B ₁+ B ₂, B ₃+ B ₄}=9.

Source

POJ Monthly--2007.07.08, Chen, Danqi

题目：http://poj.org/problem?id=324 5

题意：给你一个数列对，其实就是两个数列，你可以把数列分成任意段，要求如下：

1.对于任意元素p和元素q(p<q)在不同段中，那么有Bp>Aq

2.所有段的A元素的最大值的和要小于给定的限制

求满足上面条件下，每个段B元素的和的最大值

分析：首先考虑第一个限制，我们会发现，只有有数列后面的aj大于等于当前的bi，那么i到j的所有元素必须在一个块，那么就把他们合并，合并就是a取最大值，b取和，具体怎么合并有很多方法，我们可以将b元素的下标，按b从小到大排序，然后反向枚举a元素，标记所有小于a元素的b元素都标记上最大的标号，然后扫描一遍合并所有块

剩下的就是第二个限制了，我们会发现，第二个限制就是这题的反向求解，我们只要二分答案，用之前那题的做法判断可行性就行了

代码：

#include<set>
#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
const int mm=55555;
int a[mm],q[mm],p[mm],f[mm],b[mm];
int i,j,n,limit,l,r,m,ans;
multiset<int>sbt;
bool check(int s)
{
    int i,p,l=0,r=-1,tmp;
    __int64 sum=0;
    sbt.clear();
    for(p=i=1;i<=n;++i)
    {
        sum+=b[i];
        while(sum>s)sum-=b[p++];
        if(p>i)return 0;
        while(l<=r&&a[i]>=a[q[r]])
        {
            if(l<r)sbt.erase(f[q[r-1]]+a[q[r]]);
            --r;
        }
        q[++r]=i;
        if(l<r)sbt.insert(f[q[r-1]]+a[q[r]]);
        while(q[l]<p)
        {
            if(l<r)sbt.erase(f[q[l]]+a[q[l+1]]);
            ++l;
        }
        f[i]=f[p-1]+a[q[l]];
        tmp=*sbt.begin();
        if(l<r&&f[i]>tmp)f[i]=tmp;
    }
    return f[n]<=limit;
}
bool cmp(int u,int v)
{
    return b[u]<b[v];
}
int main()
{
    while(~scanf("%d%d",&n,&limit))
    {
        for(i=1;i<=n;++i)
            scanf("%d%d",&a[i],&b[i]),q[i]=p[i]=i;
        sort(p+1,p+n+1,cmp);
        for(j=1,i=n;i>0;--i)
            while(j<=n&&b[p[j]]<=a[i])q[p[j++]]=i;
        for(j=i=1;i<=n;i=l,++j)
        {
            a[j]=a[i],b[j]=b[i];
            for(l=i+1,r=max(q[i],i);l<=r;++l)
            {
                b[j]+=b[l];
                a[j]=max(a[j],a[l]);
                if(q[l]>r)r=q[l];
            }
        }
        n=j-1,l=0,r=2147483647;
        while(l<=r)
        {
            m=(l+r)>>1;
            if(check(m))ans=m,r=m-1;
            else l=m+1;
        }
        printf("%d\n",ans);
    }
    return 0;
}