Poj 3245 sequence partitioning (DP + binary + monotonic queue + SBT)

Sequence partitioning Time limit:8000 ms   Memory limit:65536 K Total submissions:710   Accepted:207 Case time limit:5000 Ms Description Given a sequenceNOrdered pairs of positive integers (AI,Bi), You have to partition it into several contiguous parts. LetPBe the number of these parts, whose boundaries are (L1,R1 ), (L2,R2 ),...,(Lp,RP), Which satisfyLi=RI −1 + 1,Li≤Ri,L1 = 1,RP=N. The parts themselves also satisfy the following restrictions: For any two pairs (AP,BP),(AQ,BQ), Where (AP,BP) Is belongs toTPTh part and (AQ,BQ) TheTQTh part. IfTP<TQ, ThenBP>AQ. LetMiBe the maximumA-Component of elements inITh part, say Mi= Max {Ali,ALI +1 ,...,Ari}, 1 ≤I≤P It is provided that Where limit is a given integer. LetSiBe the sumB-Components of elements inITh part. Now I want to minimize the value Max {Si: 1 ≤I≤ P} Cocould 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 {A1,A2 ,...,An} ≤Limit Output Output the minimum target value. Sample Input 4 64 33 52 52 4 Sample output 9 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. ThenB1>A3,B1>A4,B2>A3,B2>A4, Max {A1,A2} + max {A3,A4} ≤ 6, and minimum max {B1 +B2,B3 +B4} = 9. Source Poj monthly -- 2007.07.08, Chen, danqi

Question: http://poj.org/problem? Id = 3245

Question: Let's give you a pair of columns, which are actually two columns. You can divide the series into any segment. The requirements are as follows:

1. For any element P and element Q (P <q) in different segments, there are BP> AQ

2. The sum of the maximum values of element a of all segments must be smaller than the given limit.

Evaluate the maximum value of the sum of Element B in each segment that meets the preceding conditions.

Analysis: first, consider the first restriction. We will find that if the Aj after a series is greater than or equal to the current bi, all the elements of I to J must be in one block, then they will be merged. Merging means that a takes the maximum value and B takes the sum. There are many ways to merge them. We can sort the subscript of Element B in ascending order of Element B, then, element a is enumerated in reverse order, and all elements B smaller than element a are marked with the largest label, and then all parts are merged once.

The rest is the second limit. We will find that the second limit is the reverse solution to this question. We only need to have two answers and use the previous practice to determine the feasibility.

Code:

#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;}