HDU 3415

Max Sum of Max-K-sub-sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5616    Accepted Submission(s): 2024

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

 

RMQ 尝试失败。。。

单调队列，sum[i] 代表 1 ~ i 的累加值 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int MAX_N = 100005;
const int INF = 1e9;
int a[2 * MAX_N],sum[2 * MAX_N];
int N,K;
int deque[2 * MAX_N];

int cal(int x) {
        return x > N ? x - N : x;
}

void solve() {
        int anssum = -INF,anss,anse;
        int s = 0,e = 0;
        for(int i = 0; i <= 2 * N; ++i) {
                if(s != e) {
                        int t = sum[i] - sum[ deque[s]];
                        if(anssum < t || anssum == t && cal(deque[s] + 1) < anss ||
                           anssum == t && cal(deque[s] + 1) == anss && (i - deque[s]) < (anse - anss + 1)) {
                                anssum = t;
                                anss = cal(deque[s] + 1);
                                anse = cal(i);
                        }
                }

                while(s < e && sum[i] < sum[ deque[e - 1] ]) --e;
                deque[e++] = i;

                if(deque[s] == i - K) ++s;
        }

        printf("%d %d %d
",anssum,anss,anse);
}

int main()
{
    //freopen("sw.in","r",stdin);
    int t;
    scanf("%d",&t);

    while(t--) {
            scanf("%d%d",&N,&K);
            memset(sum,0,sizeof(sum));
            for(int i = 1; i <= N; ++i) {
                    scanf("%d",&a[i]);
            }
            for(int i = N + 1; i <= 2 * N; ++i) {
                    a[i] = a[i - N];
            }

            for(int i = 1;  i <= 2 * N; ++i) {
                    sum[i] += sum[i - 1] + a[i];
            }
            solve();

    }
    //cout << "Hello world!" << endl;
    return 0;
}