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Division and Union CodeForces - 1101C （排序后处理）
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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
Input
3
2
5 5
2 3
3
3 5
2 3
2 3
3
3 3
4 4
5 5
Output
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.
题意：给你N个区间，让你把这N个区间分成2个非空的集合，使不存在任意一个元素x，它即被第一个集合的某一个区间包含即L<=x<=R，也被第二个集合的某些区间包含。
如果不可以分，输出-1，如果可以，输出1~n个数，代表第i的区间放在第d个集合，d为1或2.
思路，根据L和R把区间排序后，先把排序后的第一个区间的L和R作为第一个集合的总L和R，那么我们来维护这个L和R，使第一个集合的L~R是一个连续的区间。（L~R每一个元素都可以在第一个集合中找到区间包含）
接下来从2~n遍历区间
如果下一个区间和L~R有交集，那么加入到第一个集合，更新L和R，
否则加入到第二个集合之中。
细节见代码：
#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=;
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,,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[].l;
ri=v[].r;
int is2=;
ans[v[].id]=;
for(int i=;i<=n-;i++)
{
temp=v[i];
if(temp.l<=le&&temp.r>=ri)
{
le=temp.l;
ri=temp.r;
ans[v[i].id]=;
}else if(temp.l<=ri&&temp.r<=ri)
{
// ri=temp.r;
ans[v[i].id]=;
}else if(temp.l<=ri&&temp.r>ri)
{
ri=temp.r;
ans[v[i].id]=;
}
else if(temp.l>ri)
{
is2=;
ans[v[i].id]=;
} }
if(is2)
{
repd(i,,n)
{
printf("%d ",ans[i]);
}
printf("\n");
}else
{
printf("-1\n");
}
}
return ;
} inline void getInt(int* p) {
char ch;
do {
ch = getchar();
} while (ch == ' ' || ch == '\n');
if (ch == '-') {
*p = -(getchar() - '');
while ((ch = getchar()) >= '' && ch <= '') {
*p = *p * - ch + '';
}
}
else {
*p = ch - '';
while ((ch = getchar()) >= '' && ch <= '') {
*p = *p * + ch - '';
}
}
}
