UVA 12697 Minimal Subarray Length

Minimal Subarray Length

Time Limit: 3000ms

Memory Limit: 131072KB

This problem will be judged on UVALive. Original ID: 6609
64-bit integer IO format: %lld      Java class name: Main

You are given an integer sequence of length N and another value X. You have to find a contiguous subsequence of the given sequence such that the sum is greater or equal to X. And you have to find that segment with minimal length.

Input

First line of the input file contains T the number of test cases. Each test case starts with a line containing 2 integers N (1 ≤ N ≤ 500000) and X (−10⁹ ≤ X ≤ 10⁹). Next line contains N integers denoting the elements of the sequence. These integers will be between −10⁹ to 10⁹ inclusive.

Output

For each test case output the minimum length of the sub array whose sum is greater or equal to X. If there is no such array, output ‘-1’.

Sample Input

3
5 4
1 2 1 2 1
6 -2
-5 -6 -7 -8 -9 -10
5 3
-1 1 1 1 -1

Sample Output

3
-1
3

解题：先求和。维护一个队列，下标单调，值也单调。对于i,j.如果sum[i] <= sum[j] && i > j ,那么i肯定比j好。去掉j.

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <climits>
 7 #include <vector>
 8 #include <queue>
 9 #include <cstdlib>
10 #include <string>
11 #include <set>
12 #include <stack>
13 #define LL long long
14 #define pii pair<int,int>
15 #define INF 0x3f3f3f3f
16 using namespace std;
17 const int maxn = 500100;
18 LL sum[maxn] = {0};
19 int n,x,inc[maxn];
20 int main() {
21     int t,i,lt,rt,ans;
22     scanf("%d",&t);
23     while(t--){
24         scanf("%d %d",&n,&x);
25         for(i = 1; i <= n; i++){
26             scanf("%lld",sum+i);
27             sum[i] += sum[i-1];
28         }
29         lt = 0;
30         inc[0] = rt = 1;
31         ans = n+1;
32         for(i = 2; i <= n; i++){
33             while(lt < rt && sum[i] <= sum[inc[rt-1]]) rt--;
34             inc[rt++] = i;
35             while(lt + 1 < rt && sum[i] - sum[inc[lt]] >= x){
36                 ans = min(ans,i-inc[lt]);
37                 lt++;
38             }
39         }
40         ans == n+1?puts("-1"):printf("%lld\n",ans);
41     }
42     return 0;
43 }