HDOJ 3415 Max Sum of Max-K-sub-sequence(单调队列）

因为是circle sequence,可以在序列最后+序列前n项(或前k项);利用前缀和思想，预处理出前i个数的和为sum[i],则i~j的和就为sum[j]-sum[i-1],对于每个j,取最小的sum[i-1],这就转成一道单调队列了,维护k个数的最小值。

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#include<cstdio>

#include<deque>

#define rep(i,n) for(int i=0;i<n;i++)

#define Rep(i,l,r) for(int i=l;i<=r;i++)

using namespace std;

const int maxn=100000*2+5;

const int inf=1<<30;

int sum[maxn];

deque<int> q;

deque<int> num;

int main()

{

freopen("test.in","r",stdin);

freopen("test.out","w",stdout);

int kase;

scanf("%d",&kase);

while(kase--) {

sum[0]=0;

int n,k,t;

scanf("%d%d",&n,&k);

Rep(i,1,n) {

scanf("%d",&t);

sum[i]=sum[i-1]+t;

}

Rep(i,1,n) sum[i+n]=sum[n]+sum[i];

while(!q.empty()) { q.pop_back(); num.pop_back(); }

int ans[3]={-inf,0,0};

rep(i,n+n) {

if(i && sum[i]-q.front()>ans[0]) {

ans[0]=sum[i]-q.front();

ans[1]=num.front()+1; ans[2]=i;

}

if(!q.empty()) {

if(num.front()+k<i+1) { q.pop_front(); num.pop_front(); }

while(!q.empty() && q.back()>=sum[i]) { 

   q.pop_back();

num.pop_back(); 

}

}

q.push_back(sum[i]);

num.push_back(i);

}

if(ans[2]>n) ans[2]%=n;

printf("%d %d %d\n",ans[0],ans[1],ans[2]);

}

return 0;

}

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Max Sum of Max-K-sub-sequenceTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6213    Accepted Submission(s): 2270

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

 

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