hdu3415 Max Sum of Max-K-sub-sequence 单调队列
2014-05-14 15:46
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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): 5692 Accepted Submission(s): 2060
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
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题意:给定一组环状数列,求出其中长度不超过k的数字之和的最大值,并求出对应的首位位置
思路:这个题目属于常见题型,给定一组环状数列,求出其中长度不超过k的数字之和的最大值用到了前缀和的技巧,比如要求i,j间所有数字之和,可以提前在输入的时候 计算出从0~i的数字之和及从0~j的数字之和,那么sum[i~j]=sum[j]-sum[i];现在的问题是求出最大的sum[i~j].那么我们可以固定sum[j]的值,当sum[i]最小时,整个式子值最大,显然会用到单调递增队列,不断维护进队出队操作即可,保持单调递增队列的单调性不被破坏,则队头元素始终是最小的值。详见程序:
#include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int MAXN=1000000+100; const int inf=0x3fffffff; int n,k,head,tail; int a[MAXN],sum[2*MAXN]; struct node { int val,index; }que[MAXN]; int main() { //freopen("text.txt","r",stdin); int T; scanf("%d",&T); while(T--) { scanf("%d%d",&n,&k); sum[0]=0; for(int i=1;i<=n;i++) { scanf("%d",&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]; head=tail=0; int start=1,end=1,ans=-inf; for(int i=1;i<n+k;i++) { while(head<tail && que[tail-1].val>sum[i-1]) tail--; que[tail].val=sum[i-1]; que[tail++].index=i-1; if(head<tail && que[head].index<i-k) head++; if(sum[i]-que[head].val>ans) { ans=sum[i]-que[head].val; start=que[head].index+1; end=i; } } if(end>n) end-=n; printf("%d %d %d\n", ans,start,end); } return 0; }
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