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.
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).
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
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
7 1 3
7 6 2
-1 1 1
Author
shǎ崽@HDU
Source
Recommend
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 = ;
int num[MAXN << ];
int Q[MAXN << ];
int sum[MAXN << ];
int main(){
int T, n, k;
scanf("%d", &T);
while(T--){
scanf("%d%d", &n, &k);
int head = , tail = -;
sum[] = ;
for(int i = ; i <= n; i++){
scanf("%d", num + i);
sum[i] = sum[i - ] + num[i];
}
for(int i = n + ; i <= n + k; i++){
num[i] = num[i - n];
sum[i] = sum [i - ] + num[i];
}
int ans = -0x3f3f3f3f, L = , R = ;
for(int i = ; i < n + k; i++){
while(head <= tail && sum[i - ] < sum[Q[tail]])
tail--;
while(head <= tail && i - Q[head]> k)head++;
Q[++tail] = i - ;
if(sum[i] - sum[Q[head]] > ans){
ans = sum[i] - sum[Q[head]];
L = Q[head] + ;
R = i;
}
}
printf("%d %d %d\n", ans, L>n?L-n:L, R>n?R-n:R);
}
return ;
}
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 = ;
int num[MAXN];
int main(){
int T;
int n, k;
scanf("%d", &T);
while(T--){
scanf("%d%d", &n, &k);
for(int i = ; i < n; i++){
scanf("%d", num + i);
}
for(int i = n; i < * n; i++)
num[i] = num[i - n];
int l = , r = , ans = -0x3f3f3f3f, cur = , L = , R = , cnt = ;
for(int i = ; i < * n; i++){
cur += num[i];
cnt++;
if(cur <= num[i] || cnt > k){
l = i;
cur = num[i];
cnt = ;
} r = i;
if(cur > ans){
L = l;
R = r;
ans = cur;
}
}
printf("%d %d %d\n", ans, L + > n ? L + - n: L + , R + > n ? R + - n: R + );
}
return ;
}