POJ 3415 Max Sum of Max-K-sub-sequence （线段树+dp思想）

 

 

Max Sum of Max-K-sub-sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5084    Accepted Submission(s): 1842

Problem Description

Given a circle sequence A[1],A[2],A[3]......A[n]. Circle sequence means the left neighbour of A[1] is A[n] , and the right neighbour of A[n] is A[1].
Now your job is to calculate the max sum of a Max-K-sub-sequence. Max-K-sub-sequence means a continuous non-empty sub-sequence which length not exceed K.

 

Input

The first line of the input contains an integer T(1<=T<=100) which means the number of test cases. 
Then T lines follow, each line starts with two integers N , K(1<=N<=100000 , 1<=K<=N), then N integers followed(all the integers are between -1000 and 1000).

 

Output

For each test case, you should output a line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the minimum start position, if still more than one , output the minimum length of them.

 

Sample Input

4 6 3 6 -1 2 -6 5 -5 6 4 6 -1 2 -6 5 -5 6 3 -1 2 -6 5 -5 6 6 6 -1 -1 -1 -1 -1 -1

 

Sample Output

7 1 3 7 1 3 7 6 2 -1 1 1

 

Author

shǎ崽@HDU

 

Source

HDOJ Monthly Contest – 2010.06.05

 

Recommend

lcy

 

 

题目大意：T 组数据，求 n 个数 连续子串的最大和是多少，子串要求长度不超过 k，以及这是个环形，如果多解，满足起点be最小，其次终点en最小

解题思路：枚举每个起点be，终点en一定是在 be<=en<=be+k-1 这个范围内，所以求这个范围内的连续最长和即可，可以用 sum[en] -sum[be-1] ，其中sum[x]表示前x个数的和，所以即选出 be<=en<=be+k-1这个范围内最大sum[i]，用线段树过了，但是rmq算法超时，郁闷！

线段树算法，AC

#include <iostream>
#include <cmath>
#include <cstdio>
using namespace std;

const int maxn=200010;
int d[maxn],sum[maxn],n,k,mx,mn;

struct node{
	int l,r,minc,maxc;
}a[4*maxn];

void input(){
	scanf("%d%d",&n,&k);
	for(int i=1;i<=n;i++){
		scanf("%d",&d[i]);
		d[i+n]=d[i];
	}
	sum[0]=0;
	for(int i=1;i<=2*n;i++) sum[i]=sum[i-1]+d[i];
}

void build(int l,int r,int k){
	a[k].l=l;
	a[k].r=r;
	if(l<r){
		int mid=(l+r)/2;
		build(l,mid,2*k);
		build(mid+1,r,2*k+1);
		a[k].maxc=max(a[2*k].maxc,a[2*k+1].maxc);
		a[k].minc=min(a[2*k].minc,a[2*k+1].minc);
	}else{
		a[k].maxc=sum[l];
		a[k].minc=sum[l];
	}
}

void query(int l,int r,int k){
	if(l<=a[k].l && a[k].r<=r){
		//cout<<a[k].maxc<<" "<<a[k].minc<<endl;
		mx=max(mx,a[k].maxc);
		mn=min(mn,a[k].minc);
	}else{
		int mid=(a[k].l+a[k].r)/2;
		if(r<=mid) query(l,r,2*k);
		else if(l>=mid+1) query(l,r,2*k+1);
		else{
			query(l,mid,2*k);
			query(mid+1,r,2*k+1);
		}
	}
}

void computing(){
	build(0,2*n,1);
	int ans=-(1<<30),be=1,en=1;
    for(int i=1;i<=n;i++){
    	mx=-(1<<30);
		query(i,i+k-1,1);
        if(mx-sum[i-1]>ans){
            ans=mx-sum[i-1];
            be=i;
        }
    }
    for(int i=be;i<=be+k-1;i++){
        if(sum[i]-sum[be-1]==ans){
            en=i;
            break;
        } 
    }
    printf("%d %d %d
",ans,be>n?be-n:be,en>n?en-n:en);
}

int main(){
	int t;
	scanf("%d",&t);
	while(t-- >0){
		input();
		computing();
	}
	return 0;
}

rmq算法，超时，郁闷中

#include <iostream>
#include <cmath>
#include <cstdio>
using namespace std;

const int maxn=300005*2;
int maxsum[maxn][20],minsum[maxn][20],flog[maxn],d[maxn],sum[maxn],n,k;

void ini(){
    int r=2,cnt=0;
    for(int i=1;i<maxn;i++){
        if(i<r) flog[i]=cnt;
        else{
            flog[i]=++cnt;
            r=r<<1;
        }
    }
}

void input(){
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++){
        scanf("%d",&d[i]);
        d[i+n]=d[i];
    }
}

void getrmq(){
    for(int i=1;i<=2*n;i++){
        sum[i]=sum[i-1]+d[i];
        maxsum[i][0]=sum[i];
        minsum[i][0]=sum[i];    
    }
    for(int j=1;j<=flog[2*n];j++)
    for(int i=1;i<=2*n;i++){
        if(i+(1<<j)-1<=2*n){
            maxsum[i][j]=max(maxsum[i][j-1],maxsum[i+(1<<(j-1))][j-1]);
            minsum[i][j]=min(minsum[i][j-1],minsum[i+(1<<(j-1))][j-1]);
        }
    }
}

void computing(){
    getrmq();
    int ans=-(1<<30),be=1,en=1,l,r,x,tmp;
    for(int i=1;i<=n;i++){
        l=i,r=i-1+k,x=flog[r-l+1];
        tmp=max(maxsum[l][x],maxsum[r-(1<<x)+1][x]);
        if(tmp-sum[i-1]>ans){
            ans=tmp-sum[i-1];
            be=i;
        }
    }
    for(int i=be;i<=be+k-1;i++){
        if(sum[i]-sum[be-1]==ans){
            en=i;
            break;
        } 
    }
    printf("%d %d %d
",ans,be>n?be-n:be,en>n?en-n:en);
}

int main(){
    ini();
    int t;
    scanf("%d",&t);
    while(t-- >0){
        input();
        computing();
    }
    return 0;
}