C2. Make Nonzero Sum (hard version)

C2. Make Nonzero Sum (hard version) time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output

This is the hard version of the problem. The difference is that in this version the array contains zeros. You can make hacks only if both versions of the problem are solved.

You are given an array [a1,a2,…an][a1,a2,…an] consisting of integers −1−1, 00 and 11. You have to build a partition of this array into the set of segments [l1,r1],[l2,r2],…,[lk,rk][l1,r1],[l2,r2],…,[lk,rk] with the following property:

• Denote the alternating sum of all elements of the ii-th segment as sisi: sisi = ali−ali+1+ali+2−ali+3+…±ariali−ali+1+ali+2−ali+3+…±ari. For example, the alternating sum of elements of segment [2,4][2,4] in array [1,0,−1,1,1][1,0,−1,1,1] equals to 0−(−1)+1=20−(−1)+1=2.
• The sum of sisi over all segments of partition should be equal to zero.

Note that each sisi does not have to be equal to zero, this property is about sum of sisi over all segments of partition.

The set of segments [l1,r1],[l2,r2],…,[lk,rk][l1,r1],[l2,r2],…,[lk,rk] is called a partition of the array aa of length nn if 1=l1≤r1,l2≤r2,…,lk≤rk=n1=l1≤r1,l2≤r2,…,lk≤rk=n and ri+1=li+1ri+1=li+1 for all i=1,2,…k−1i=1,2,…k−1. In other words, each element of the array must belong to exactly one segment.

You have to build a partition of the given array with properties described above or determine that such partition does not exist.

Note that it is not required to minimize the number of segments in the partition.

Each test contains multiple test cases. The first line contains the number of test cases tt (1≤t≤100001≤t≤10000). Description of the test cases follows.

The first line of each test case contains an integer nn (1≤n≤2000001≤n≤200000) — the length of array aa.

The second line of each test case contains nn integers a1,a2,…,ana1,a2,…,an (aiai is −1−1, 00, or 11) — the elements of the given array.

It's guaranteed that the sum of nn over all test cases does not exceed 200000200000.

For each test case print an integer kk — the number of segments in the partition. If required partition does not exist, print −1−1.

If partition exists, in the ii-th of the following kk lines print two integers lili and riri — description of the ii-th segment. The following conditions should be satisfied:

• li≤rili≤ri for each ii from 11 to kk.
• li+1=ri+1li+1=ri+1 for each ii from 11 to (k−1)(k−1).
• l1=1,rk=nl1=1,rk=n.

If there are multiple correct partitions of the array, print any of them.

inputCopy
5
4
0 0 0 0
7
-1 1 0 1 0 1 0
5
0 -1 1 0 1
3
1 0 1
1
1
outputCopy
4
1 1
2 2
3 3
4 4
4
1 1
2 2
3 5
6 7
-1
2
1 1
2 3
-1

#include <iostream>
#include <cstring>
using namespace std;
const int N=2e5+10;
int t,n;
int a[N],sum;
bool p[N];
typedef struct{
    int l,r;
}node;
node res[N];
int main(){
    cin>>t;
    while(t--){
        cin>>n;
        sum=0;
        memset(p,0,sizeof p);
        for(int i=1;i<=n;i++) cin>>a[i],sum+=a[i];
        for(int i=2;i<=n;i++){
            if(sum>0&&a[i]==1){
                p[i-1]=1;
                sum-=2;
                i++;
            }
            else if(sum<0&&a[i]==-1){
                p[i-1]=1;
                sum+=2;
                i++;
            }
        }
        if(sum!=0){
            cout<<-1<<endl;
            continue;
        }
        int l,r,num=0;
        for(int i=1;i<=n;i++){
            l=i;
            r=i;
            if(p[i]){
            while(i<=n&&p[i]) r=i+1,i+=2;
            i--;
        }
            res[num++]={l,r};
        }
        cout<<num<<endl;
        for(int i=0;i<num;i++) cout<<res[i].l<<" "<<res[i].r<<endl;
    }
}