HDU 3415 Max Sum of Max-K-sub-sequence [单调队列]【杂类】
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题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3415
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Max Sum of Max-K-sub-sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7522 Accepted Submission(s): 2776
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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题目大意:
给你一个成环的数列,让你寻找其中区间长度不大于k的区间最大和与 区间的起始位置和结束位置。
解题思路:
因为做的题目就是单调队列的专题 就没想算法
然后考虑成环的 所以要把区间变成2倍的 这样就有环了,,,
然后因为要取区间的和 而维护单调队列的时候也只是维护区间的左界 为了方便计算区间的值 所以把数组变成前缀和的形式
然后在维护的时候个人维护的是左界限的左一位的值 计算区间和比较容易
总的来说就是
枚举右界限 维护一下每次最优的左界线 然后把不符合结果的出队列就好了
附本题代码
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#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>typedef long long int LL ;using namespace std;#define Rep(i,a,b) for(int (i)=(a);(i)<=(b);(i)++)const int N = 2e6+5;int a[N];int p[N]; //单调队列 维护左界线下标int main(){ int _; scanf("%d",&_); while(_--){ int n,k; scanf("%d%d",&n,&k); a[0]=0; Rep(i,1,n) scanf("%d",&a[i]),a[i+n]=a[i]; n+=k-1; //成环的话 因为最大的区间是K 座椅长度只需要n+k-1就行了 Rep(i,2,n) a[i]+=a[i-1]; int l=1,r=0,ml=-1,mr=-1,mx=-0x3f3f3f3f,tem=0; p[0]=p[++r]=0;//其实P[0] 是没有用的。。 Rep(i,1,n){ while(l<=r && p[l] < i-k) l++; //不满足K的需要出队列 tem = a[i]-a[p[l]];// 一定要在更新之前计算 否则结果可能因为负值出现错误 if( tem > mx ) ml=p[l]+1,mr=i,mx=tem; while(l<=r && a[i] < a[ p[r] ]) r--;//更新最优解 p[++r]=i; } n-=k-1; mr=(mr>n?mr-n:mr);//大于n的位置 处理一下 printf("%d %d %d\n",mx,ml,mr); } return 0;}
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