14.Binary Deque
题面翻译
Binary Deque - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)
有多组数据。
每组数据给出 \(n\) 个数，每个数为 \(0\) 或 \(1\) 。你可以选择从两边删数，求至少删几个数才可以使剩下的数总和为 \(s\) 。
如果不能达到 \(s\) ，则输出 \(-1\) 。
题目描述
Slavic has an array of length $ n $ consisting only of zeroes and ones. In one operation, he removes either the first or the last element of the array.
What is the minimum number of operations Slavic has to perform such that the total sum of the array is equal to $ s $ after performing all the operations? In case the sum $ s $ can't be obtained after any amount of operations, you should output -1.
输入格式
The first line contains a single integer $ t $ ( $ 1 \leq t \leq 10^4 $ ) — the number of test cases.
The first line of each test case contains two integers $ n $ and $ s $ ( $ 1 \leq n, s \leq 2 \cdot 10^5 $ ) — the length of the array and the needed sum of elements.
The second line of each test case contains $ n $ integers $ a_i $ ( $ 0 \leq a_i \leq 1 $ ) — the elements of the array.
It is guaranteed that the sum of $ n $ over all test cases doesn't exceed $ 2 \cdot 10^5 $ .
输出格式
For each test case, output a single integer — the minimum amount of operations required to have the total sum of the array equal to $ s $ , or -1 if obtaining an array with sum $ s $ isn't possible.
样例 #1
样例输入 #1
7
3 1
1 0 0
3 1
1 1 0
9 3
0 1 0 1 1 1 0 0 1
6 4
1 1 1 1 1 1
5 1
0 0 1 1 0
16 2
1 1 0 0 1 0 0 1 1 0 0 0 0 0 1 1
6 3
1 0 1 0 0 0
样例输出 #1
0
1
3
2
2
7
-1
提示
In the first test case, the sum of the whole array is $ 1 $ from the beginning, so we don't have to make any operations.
In the second test case, the sum of the array is $ 2 $ and we want it to be equal to $ 1 $ , so we should remove the first element. The array turns into $ [1, 0] $ , which has a sum equal to $ 1 $ .
In the third test case, the sum of the array is $ 5 $ and we need it to be $ 3 $ . We can obtain such a sum by removing the first two elements and the last element, doing a total of three operations. The array turns into $ [0, 1, 1, 1, 0, 0] $ , which has a sum equal to $ 3 $ .
思路：
本题可以使用双指针来动态获取最优解，我们可以先计算出所有元素之和sum，若sum小于s，则不可能达到目标，输出-1并接着处理下一个样例，若s==sum，则正好操作次数为0，可以输出0并直接返回。否则，我们使用双指针的思路，此处为快慢指针，我们维护一个动态区间。
- 若这个动态区间内的元素之和sum=s，则我们此时更新答案，并让快指针向右走
- 若这个动态区间内的元素之和 sum<s，则我们需要扩大区间范围，此时让快指针r接着向右走
- 若这个动态区间内的元素之和sum>s，则我们需要缩小区间范围，找到最优的答案，此时应该是让慢指针向右走，逐步缩小这个区间的大小。
需要注意的是，sum<s和sum>s时，对l和r做自增运算的顺序不同，sum>s时，我们需要加上r右边的这个元素，所以应该是让r先自增，再让sum+=a[r]
而sum<s时，我们要减去l这个位置的元素，则我们是先sum-=a[l]，再让l++。
代码：
#include<bits/stdc++.h>
using namespace std;
const int N=2e5+10;
int n,s;
int a[N];
void solve(){
int n,s;
cin>>n>>s;
int sum=0;
//先计算所有的元素之和，若所有元素之和还小于s，则不可能达到目标，若刚好等于s，则操作次数为0次。
for(int i=1;i<=n;i++){
cin>>a[i];
sum+=a[i];
}
if(sum<s){
cout<<-1<<"\n";
return;
}
if(sum==s){
cout<<0<<'\n';
return;
}
sum=a[1];
int l=1,r=1,ans=1e10;
while(r<=n&&l<=r){
//若当前区间的和等于s，则此时更新答案，并且让快指针接着往后走
if(sum==s){
//当前所需要删除的数的个数为：总个数n-区间的长度。
ans=min(ans,n-(r-l+1));
r++;
sum+=a[r];
}
//若当前区间的和小于s，则让快指针向右走，扩大区间的范围。
if(sum<s){
r++;
sum+=a[r];
}
//若当前区间的和大于s，则让慢指针向右走，减少区间的长度，看看能不能达到最优的答案，
//注意这里是先减去a[l],再让l++，与上面不同。
if(sum>s){
sum-=a[l];
l++;
}
}
cout<<ans<<'\n';
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
int t=1;cin>>t;
while(t--) solve();
return 0;
}