qscoj#19D(单调队列)

题目链接：http://qscoj.cn/problem/130/

题意：中文题诶～

思路：直接用单调栈搞一下就好了

代码：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3
 4 const int MAXN=1e6+10;
 5 const int inf=1e9;
 6 int sum[MAXN], q[MAXN];
 7
 8 int main(void){
 9     int n, k;
10     while(scanf("%d%d", &n, &k)!=EOF){
11         for(int i=1; i<=n; i++){
12             int x;
13             scanf("%d", &x);
14             sum[i]=sum[i-1]+x;
15         }
16         sum[n+1]=-inf;
17         int ans=0, front=0, rear=0;
18         for(int i=1; i<=n+1; i++){
19             ans=max(ans, sum[i]);
20             if((front==rear||sum[i]>sum[q[front]])){
21                 if(i-q[rear+1]+1<k){
22                     q[++front]=i;
23                 }else{
24                     while(i-q[rear+1]+1>=k&&front>rear){
25                         rear++;
26                     }
27                     q[++front]=i;
28                 }
29             }else if(sum[i]<sum[q[front]]){
30                 while(front>rear&&sum[i]<sum[q[front]]){
31                     ans=max(ans, sum[q[front]]-sum[q[rear+1]]);
32                     front--;
33                 }
34                 while(i-q[rear+1]+1>=k&&front>rear){
35                     rear++;
36                 }
37                 q[++front]=i;
38             }
39         }
40         printf("%d\n", ans);
41     }
42     return 0;
43 }