Max Sum of Max-K-sub-sequence(单调队列)

Max Sum of Max-K-sub-sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7034 Accepted Submission(s): 2589

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 | We have carefully selected several similar problems for you: 3423 3417 3418 3419 3421

题解：让找距离小于等于k的连续一段区间的值最大，数据是环状的；

要用单调队列；本来想着贪心，果断wa，单调队列注意，找到前缀和，现在只需要根据r找l，l在单调队列里找，单调队列要保证从小到大。这样就保证了队头是最小值，注意单调队列放的是i(初态)，所以要放i - 1；这点错了好久。当前的位置是终态，也就是i

代码：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN = 1000100;
int num[MAXN << 1];
int Q[MAXN << 1];
int sum[MAXN << 1];
int main(){
    int T, n, k;
    scanf("%d", &T);
    while(T--){
        scanf("%d%d", &n, &k);
        int head = 0, tail = -1;
        sum[0] = 0;
        for(int i = 1; i <= n; i++){
            scanf("%d", num + i);
            sum[i] = sum[i - 1] + num[i];
        }
        for(int i = n + 1; i <= n + k; i++){
            num[i] = num[i - n];
            sum[i] = sum [i - 1] + num[i];
        }
        int ans = -0x3f3f3f3f, L = 0, R = 0;
        for(int i = 1; i < n + k; i++){
            while(head <= tail && sum[i - 1] < sum[Q[tail]])
                tail--;
            while(head <= tail && i - Q[head]> k)head++;
            Q[++tail] = i - 1;
            if(sum[i] - sum[Q[head]] > ans){
                ans = sum[i] - sum[Q[head]];
                L = Q[head] + 1;
                R = i;
            }
        }
        printf("%d %d %d
", ans, L>n?L-n:L, R>n?R-n:R);
    }
    return 0;
}

贪心wa；由于规定了最大长度，这样贪心就不行了；

wa代码 :

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN = 1000010;
int num[MAXN];
int main(){
    int T;
    int n, k;
    scanf("%d", &T);
    while(T--){
        scanf("%d%d", &n, &k);
        for(int i =0 ; i < n; i++){
            scanf("%d", num + i);
        }
        for(int i = n; i < 2 * n; i++)
            num[i] = num[i - n];
            int l = 0, r = 0, ans = -0x3f3f3f3f, cur = 0, L = 0, R = 0, cnt = 0;
        for(int i = 0; i < 2 * n; i++){
            cur += num[i];
            cnt++;
            if(cur <= num[i] || cnt > k){
                l = i;
                cur = num[i];
                cnt = 1;
            }

            r = i;
            if(cur > ans){
                L = l;
                R = r;
                ans = cur;
            }
        }
        printf("%d %d %d
", ans, L + 1 > n ? L + 1 - n: L + 1, R + 1 > n ? R + 1 - n: R + 1);
    }
    return 0;
}