【hdu3415】【单调队列 】Max Sum of Max-K-sub-sequence【求长度不大于k的区间最大子串和】
2016-11-03 20:22
411 查看
传送门:http://acm.hdu.edu.cn/showproblem.php?pid=3415
题意:给出一个有N个数字(N<=10^5)的环状序列,让你求一个和最大的连续子序列(子串)。这个连续子序列的长度小于等于K。
思路:
因为序列是环状的,所以可以在序列后面复制前k-1个数字。
如果用s[i]来表示复制过后的序列的前i个数的和,那么任意一个子序列[i..j]的和就等于s[j]-s[i-1]。
对于每一个j,用s[j]减去最小的一个s[i](i>=j-k)就可以得到以j为终点长度不大于k的和最大的序列了。将原问题转化为这样一个问题后,就可以用单调队列解决了。
单调队列即保持队列中的元素单调递增(或递减)的这样一个队列,可以从两头删除,只能从队尾插入。单调队列的具体作用在于,由于保持队列中的元素满足单调性,
对于上述问题中的每个j,可以用O(1)的时间找到对应的s[i]。(保持队列中的元素单调递增的话,队首元素便是所要的元素了)。
维护方法:对于每个j,我们插入s[j-1]的下标,插入时从队尾插入。为了保证队列的单调性,我们从队尾开始删除元素,直到队尾元素对应的值比当前需要插入的s[j-1]小,
就将当前元素下标插入到队尾。之所以可以将之前的队列尾部元素全部删除,是因为它们已经不可能成为最优的元素了,因为当前要插入的元素位置比它们靠前,
对应的值比它们小。我们要找的,是满足(i>=j-k)的i中最小的s[i]。在插入元素后,从队首开始,将不符合限制条件(i<j-k)的元素全部删除,此时队列一定不为空。(因为刚刚插入了一个一定符合条件的元素)。
代码:
#include <bits/stdc++.h>
using namespace std;
#define rep(i,k,n) for(int i=k;i<=n;i++)
template<class T> void read(T&num) {
char CH; bool F=false;
for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
F && (num=-num);
}
const int inf=0x3f3f3f3f;
const int N=1e5+10;
int a
, sum[2 * N];
int n, k;
int main(){
int T;
read(T);
while(T--){
read(n), read(k);
rep(i, 1, n)
read(a[i]), sum[i] = sum[i - 1] + a[i];
for(int i = n + 1; i < n + k; i++)
sum[i] = sum[i - 1] + a[i - n];
int tmp = n;
n = n + k - 1;
deque<int>deq; int s, t, ans = -inf;
deq.clear();
rep(i, 1, n){
while(!deq.empty() && sum[deq.back()] > sum[i - 1])//不是>=,为了保留the minimum start position
deq.pop_back(); //保持队列的单调性
while(!deq.empty() && deq.front() < i - k)
deq.pop_front(); //超过k的长度则消除队列前面的元素
deq.push_back(i - 1);
if(sum[i] - sum[deq.front()] > ans){//记录,sum
-sum[m]所得出的是m+1到n之间的和
ans = sum[i] - sum[deq.front()];
s = deq.front() + 1;
t = i;
}
}
if(t > tmp)t %= tmp;
printf("%d %d %d\n", ans, s, t);
}
return 0;
}
描述:
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7535 Accepted Submission(s): 2781
Problem Description
Given a circle sequence A[1],A[2],A[3]......A
. Circle sequence means the left neighbour of A[1] is A
, and the right neighbour of A
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
题意:给出一个有N个数字(N<=10^5)的环状序列,让你求一个和最大的连续子序列(子串)。这个连续子序列的长度小于等于K。
思路:
因为序列是环状的,所以可以在序列后面复制前k-1个数字。
如果用s[i]来表示复制过后的序列的前i个数的和,那么任意一个子序列[i..j]的和就等于s[j]-s[i-1]。
对于每一个j,用s[j]减去最小的一个s[i](i>=j-k)就可以得到以j为终点长度不大于k的和最大的序列了。将原问题转化为这样一个问题后,就可以用单调队列解决了。
单调队列即保持队列中的元素单调递增(或递减)的这样一个队列,可以从两头删除,只能从队尾插入。单调队列的具体作用在于,由于保持队列中的元素满足单调性,
对于上述问题中的每个j,可以用O(1)的时间找到对应的s[i]。(保持队列中的元素单调递增的话,队首元素便是所要的元素了)。
维护方法:对于每个j,我们插入s[j-1]的下标,插入时从队尾插入。为了保证队列的单调性,我们从队尾开始删除元素,直到队尾元素对应的值比当前需要插入的s[j-1]小,
就将当前元素下标插入到队尾。之所以可以将之前的队列尾部元素全部删除,是因为它们已经不可能成为最优的元素了,因为当前要插入的元素位置比它们靠前,
对应的值比它们小。我们要找的,是满足(i>=j-k)的i中最小的s[i]。在插入元素后,从队首开始,将不符合限制条件(i<j-k)的元素全部删除,此时队列一定不为空。(因为刚刚插入了一个一定符合条件的元素)。
代码:
#include <bits/stdc++.h>
using namespace std;
#define rep(i,k,n) for(int i=k;i<=n;i++)
template<class T> void read(T&num) {
char CH; bool F=false;
for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
F && (num=-num);
}
const int inf=0x3f3f3f3f;
const int N=1e5+10;
int a
, sum[2 * N];
int n, k;
int main(){
int T;
read(T);
while(T--){
read(n), read(k);
rep(i, 1, n)
read(a[i]), sum[i] = sum[i - 1] + a[i];
for(int i = n + 1; i < n + k; i++)
sum[i] = sum[i - 1] + a[i - n];
int tmp = n;
n = n + k - 1;
deque<int>deq; int s, t, ans = -inf;
deq.clear();
rep(i, 1, n){
while(!deq.empty() && sum[deq.back()] > sum[i - 1])//不是>=,为了保留the minimum start position
deq.pop_back(); //保持队列的单调性
while(!deq.empty() && deq.front() < i - k)
deq.pop_front(); //超过k的长度则消除队列前面的元素
deq.push_back(i - 1);
if(sum[i] - sum[deq.front()] > ans){//记录,sum
-sum[m]所得出的是m+1到n之间的和
ans = sum[i] - sum[deq.front()];
s = deq.front() + 1;
t = i;
}
}
if(t > tmp)t %= tmp;
printf("%d %d %d\n", ans, s, t);
}
return 0;
}
描述:
Max Sum of Max-K-sub-sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 7535 Accepted Submission(s): 2781
Problem Description
Given a circle sequence A[1],A[2],A[3]......A
. Circle sequence means the left neighbour of A[1] is A
, and the right neighbour of A
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
相关文章推荐
- hdu3415 Max Sum of Max-K-sub-sequence(单调队列求n个数中和的最大值)
- HDOJ 题目3415 Max Sum of Max-K-sub-sequence(单调队列求区间和最大值)
- HDU 3415 Max Sum of Max-K-sub-sequence(长度不超过k的最大连续子序列和,单调队列)
- 单调队列的一个应用——求解连续区间最大值(HDU Max Sum of Max-K-sub-sequence)
- HDU 3415 Max Sum of Max-K-sub-sequence(单调队列+最大连续子串和)
- hdu3415 Max Sum of Max-K-sub-sequence 单调队列
- HDU3415 Max Sum of Max-K-sub-sequence (DP+单调队列)
- HDU3415:Max Sum of Max-K-sub-sequence(单调队列)
- hdu3415 Max Sum of Max-K-sub-sequence 单调队列
- HDU3415 - Max Sum of Max-K-sub-sequence - 单调队列
- HDU 3415 Max Sum of Max-K-sub-sequence(求长度不超过K的最大区间和)
- HDU3415:Max Sum of Max-K-sub-sequence(单调队列)
- hdu3415 Max Sum of Max-K-sub-sequence 单调队列
- 【DP+单调队列】 hdu3415 Max Sum of Max-K-sub-sequence
- hdu 3415 Max Sum of Max-K-sub-sequence 单调队列 求连续l(1<=l<=k)个数的和的最大值 数列可循环
- hdu 3415 Max Sum of Max-K-sub-sequence 单调队列dp
- hdu Max Sum of Max-K-sub-sequence 单调队列优化DP
- hdu 3415 Max Sum of Max-K-sub-sequence(单调队列)
- HDU 3415 Max Sum of Max-K-sub-sequence[单调队列优化dp]
- HDU-3415 Max Sum of Max-K-sub-sequence 单调队列