hdu 3415 Max Sum of Max-K-sub-sequence(单调队列)
2012-08-18 13:37
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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): 3728 Accepted Submission(s): 1317
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
题目:http://acm.hdu.edu.cn/showproblem.php?pid=3415
题意:给你一个环形的数列,问你连续长度为k的数的和的最大值,并且输出区间坐标,使坐标字典序最小
分析:把环形转换成直线处理即可,求出累加和数组,用优先队列维护
PS:这回总是是1Y了,就是这种感觉^_^
代码:
#include<cstdio> #include<iostream> using namespace std; const int mm=222222; int a[mm],q[mm]; int main() { int i,l,r,n,k,t,tmp,ans,u,v; scanf("%d",&t); while(t--) { scanf("%d%d",&n,&k); ans=-2e9; for(i=1;i<=n;++i) { scanf("%d",&a[i]); a[i+n]=a[i]; if(a[i]>ans) ans=a[i],u=i-1,v=i; } a[0]=0; for(i=1;i<n*2;++i) a[i]+=a[i-1]; q[l=r=0]=0; for(i=1;i<n*2;++i) { while(l<=r&&a[q[r]]>a[i])--r; q[++r]=i; while(i-q[l]>k)++l; if(q[l]<i) { tmp=a[i]-a[q[l]]; if(tmp==ans&&q[l]+1<u) u=q[l],v=i; if(tmp>ans) { ans=tmp; u=q[l],v=i; } } } printf("%d %d %d\n",ans,(u%n)+1,((v-1)%n)+1); } return 0; }
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