There are nn segments [li,ri][li,ri] for 1≤i≤n1≤i≤n. You should divide all segments into two non-empty groups in such way that there is no pair of segments from different groups which have at least one common point, or say that it's impossible to do it. Each segment should belong to exactly one group.
To optimize testing process you will be given multitest.
Input
The first line contains one integer TT (1≤T≤500001≤T≤50000) — the number of queries. Each query contains description of the set of segments. Queries are independent.
First line of each query contains single integer nn (2≤n≤1052≤n≤105) — number of segments. It is guaranteed that ∑n∑n over all queries does not exceed 105105.
The next nn lines contains two integers lili, riri per line (1≤li≤ri≤2⋅1051≤li≤ri≤2⋅105) — the ii-th segment.
Output
For each query print nn integers t1,t2,…,tnt1,t2,…,tn (ti∈{1,2}ti∈{1,2}) — for each segment (in the same order as in the input) titi equals 11 if the ii-th segment will belongs to the first group and 22 otherwise.
If there are multiple answers, you can print any of them. If there is no answer, print −1−1.
Example
3 2 5 5 2 3 3 3 5 2 3 2 3 3 3 3 4 4 5 5
2 1 -1 1 1 2
Note
In the first query the first and the second segments should be in different groups, but exact numbers don't matter.
In the second query the third segment intersects with the first and the second segments, so they should be in the same group, but then the other group becomes empty, so answer is −1−1.
In the third query we can distribute segments in any way that makes groups non-empty, so any answer of 66 possible is correct.
題意：給你N個區間，讓你把這N個區間分成2個非空的集合，使不存在任意一個元素x，它即被第一個集合的某一個區間包含即L<=x<=R，也被第二個集合的某些區間包含。
如果不可以分，輸出-1，如果可以，輸出1~n個數，代表第i的區間放在第d個集合，d為1或2.
思路，根據L和R把區間排序后，先把排序后的第一個區間的L和R作為第一個集合的總L和R，那么我們來維護這個L和R，使第一個集合的L~R是一個連續的區間。（L~R每一個元素都可以在第一個集合中找到區間包含）
接下來從2~n遍歷區間
如果下一個區間和L~R有交集，那么加入到第一個集合，更新L和R，
否則加入到第二個集合之中。
細節見代碼：
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <queue> #include <stack> #include <map> #include <set> #include <vector> #define rep(i,x,n) for(int i=x;i<n;i++) #define repd(i,x,n) for(int i=x;i<=n;i++) #define pii pair<int,int> #define pll pair<long long ,long long> #define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0) #define MS0(X) memset((X), 0, sizeof((X))) #define MSC0(X) memset((X), '\0', sizeof((X))) #define pb push_back #define mp make_pair #define fi first #define se second #define gg(x) getInt(&x) using namespace std; typedef long long ll; inline void getInt(int* p); const int maxn=1000010; const int inf=0x3f3f3f3f; /*** TEMPLATE CODE * * STARTS HERE ***/ int t; struct node { int l; int r; int id; }; typedef struct node node; std::vector<node> v; int n; bool cmp(node a,node b) { if(a.l!=b.l) { return a.l<b.l; }else { return a.r<b.r; } } int ans[maxn]; int main() { scanf("%d",&t); while(t--) { v.clear(); scanf("%d",&n); node temp; repd(i,1,n) { scanf("%d %d",&temp.l,&temp.r); temp.id=i; v.push_back(temp); } sort(v.begin(), v.end(),cmp); int le,ri; le=v[0].l; ri=v[0].r; int is2=0; ans[v[0].id]=1; for(int i=1;i<=n-1;i++) { temp=v[i]; if(temp.l<=le&&temp.r>=ri) { le=temp.l; ri=temp.r; ans[v[i].id]=1; }else if(temp.l<=ri&&temp.r<=ri) { // ri=temp.r; ans[v[i].id]=1; }else if(temp.l<=ri&&temp.r>ri) { ri=temp.r; ans[v[i].id]=1; } else if(temp.l>ri) { is2=1; ans[v[i].id]=2; } } if(is2) { repd(i,1,n) { printf("%d ",ans[i]); } printf("\n"); }else { printf("-1\n"); } } return 0; } inline void getInt(int* p) { char ch; do { ch = getchar(); } while (ch == ' ' || ch == '\n'); if (ch == '-') { *p = -(getchar() - '0'); while ((ch = getchar()) >= '0' && ch <= '9') { *p = *p * 10 - ch + '0'; } } else { *p = ch - '0'; while ((ch = getchar()) >= '0' && ch <= '9') { *p = *p * 10 + ch - '0'; } } }