HDU 3415 Max Sum of Max-K-sub-sequence(单调队列)
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Given a circle sequence A1 ,A2 ,A3 ......An . Circle sequence means the left neighbour of A1 is An , and the right neighbour of An is A1 .
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.
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.
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).
46 36 -1 2 -6 5 -56 46 -1 2 -6 5 -56 3-1 2 -6 5 -5 66 6-1 -1 -1 -1 -1 -1
【题意】一个长度为n的循环序列,在其中找一个长度为k(k<=n)的子序列,要求这个子序列的和尽可能大,最后输出最大值,并且输出子序列的起点和终点。
【分析】先用sum数组来保存前i个数的和,记得还要保存i到n+k的数的和,因为这是要构成首尾相连。然后从头开始,找子序列的最大值,双端单调队列队首永远是序列和最大的,往后依次递减,每移动一次保存一下和前一次ans的最小值,最后输出。
【用时】452 ms!!
【AC代码】
#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<queue>using namespace std;const int N=2e5+10;const int INF=1e7;int a[N],sum[N];int head;int tail;int str[N];int n,nn,k,maxn;int s,e;int min(int x,int y){ return x>y?y:x;}void push_up(int i){ str[tail++]=i;}bool isempty(){ return head==tail;}int Front(){ return str[head];}void Pop_back(){ tail--;}void Pop_front(){ head++;}void solve(){ head=tail=0; int i; for(i=1;i<=n;i++) { while(!isempty() && sum[i-1]<sum[str[tail-1]]) Pop_back(); while(!isempty() && str[head]+k<i) Pop_front(); push_up(i-1); if(sum[i]-sum[str[head]]>maxn) { maxn = sum[i]-sum[str[head]]; s = str[head]+1; e= i ; } } if(e>nn) e%=nn; printf("%d %d %d\n",maxn,s,e);}int main(){ int t; scanf("%d",&t); while(t--) { scanf("%d%d",&n,&k); nn=n; int i; for(i=1,sum[0]=0;i<=n;i++) { scanf("%d",&a[i]); sum[i]=sum[i-1]+a[i]; } for(;i<n+k;i++) sum[i]=sum[i-1]+a[i-n]; n=n+k-1; maxn=-INF; solve(); } return 0;}
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