因为是circle sequence,可以在序列最后+序列前n项(或前k项);利用前缀和思想，预处理出前i个数的和为sum[i],则i~j的和就为sum[j]-sum[i-1],对于每个j,取最小的sum[i-1],这就转成一道单调队列了,维护k个数的最小值。
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#include<cstdio>
#include<deque>
#define rep(i,n) for(int i=0;i<n;i++)
#define Rep(i,l,r) for(int i=l;i<=r;i++)
using namespace std;
const int maxn=100000*2+5;
const int inf=1<<30;
int sum[maxn];
deque<int> q;
deque<int> num;
int main()
{
freopen("test.in","r",stdin);
freopen("test.out","w",stdout);
int kase;
scanf("%d",&kase);
while(kase--) {
sum[0]=0;
int n,k,t;
scanf("%d%d",&n,&k);
Rep(i,1,n) {
scanf("%d",&t);
sum[i]=sum[i-1]+t;
}
Rep(i,1,n) sum[i+n]=sum[n]+sum[i];
while(!q.empty()) { q.pop_back(); num.pop_back(); }
int ans[3]={-inf,0,0};
rep(i,n+n) {
if(i && sum[i]-q.front()>ans[0]) {
ans[0]=sum[i]-q.front();
ans[1]=num.front()+1; ans[2]=i;
}
if(!q.empty()) {
if(num.front()+k<i+1) { q.pop_front(); num.pop_front(); }
while(!q.empty() && q.back()>=sum[i]) {
q.pop_back();
num.pop_back();
}
}
q.push_back(sum[i]);
num.push_back(i);
}
if(ans[2]>n) ans[2]%=n;
printf("%d %d %d\n",ans[0],ans[1],ans[2]);
}
return 0;
}
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Max Sum of Max-K-sub-sequenceTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6213 Accepted Submission(s): 2270
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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