題干：

DreamGrid has just found an undirected simple graph with??vertices and no edges (that's to say, it's a graph with??isolated vertices) in his right pocket, where the vertices are numbered from 1 to?. Now he would like to perform??operations of the following two types on the graph:

• 1?a?b?-- Connect the?-th vertex and the?-th vertex by an edge. It's guaranteed that before this operation, there does not exist an edge which connects vertex??and??directly.
• 2?k?-- Find the answer for the query: What's the minimum and maximum possible number of connected components after adding??new edges to the graph. Note that after adding the??edges, the graph must still be a simple graph, and the query does NOT modify the graph.

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Please help DreamGrid find the answer for each operation of the second type. Recall that a simple graph is a graph with no self loops or multiple edges.

Input

There are multiple test cases. The first line of the input is an integer?, indicating the number of test cases. For each test case:

The first line contains two integers??and??(,?), indicating the number of vertices and the number of operations.

For the following??lines, the?-th line first contains an integer??(), indicating the type of the?-th operation.

• If?, two integers??and??follow (,?), indicating an operation of the first type. It's guaranteed that before this operation, there does not exist an edge which connects vertex??and??directly.
• If?, one integer??follows (), indicating an operation of the second type. It's guaranteed that after adding??edges to the graph, the graph is still possible to be a simple graph.

?

It's guaranteed that the sum of??in all test cases will not exceed?, and the sum of??in all test cases will not exceed?.

Output

For each operation of the second type output one line containing two integers separated by a space, indicating the minimum and maximum possible number of connected components in this query.

Sample Input

1 5 5 1 1 2 2 1 1 1 3 2 1 2 3

Sample Output

3 3 2 3 1 2

解題報(bào)告：

以連通塊中元素個(gè)數(shù)為權(quán)值，建立線段樹，維護(hù)三個(gè)變量，分別是cnt（這樣的連通塊的數(shù)量），sum（連通塊的總的點(diǎn)數(shù)），pfh（各連通塊之間點(diǎn)數(shù)的平方和）（注意不是和的平方）

cnt代表連通塊數(shù)量，tot_ret代表當(dāng)前狀態(tài)下的圖，在不增長連通塊個(gè)數(shù)的前提下，可以加的邊數(shù)。

所以對于最小值很顯然。

對于最大值，首先減去塊內(nèi)連的邊，然后去線段樹查詢剩下的邊怎么加，首先填連通塊元素大小大的，這樣也就是類似查詢第k大，查到葉子結(jié)點(diǎn)，再判斷 塊中元素為tree[cur].l的 塊 我需要多少個(gè) 返回回去，就行了。

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define F first #define S second #define ll long long #define pb push_back #define pm make_pair using namespace std; typedef pair<int,int> PII; const int MAX = 2e5 + 5; int n,q,cnt,f[MAX]; ll tot_ret,ret[MAX],num[MAX]; int getf(int v) {return f[v] == v ? v : f[v] = getf(f[v]); } struct TREE {int l,r;ll cnt,sum;ll pfh; } tree[MAX<<2]; void pushup(int cur) {tree[cur].cnt = tree[cur*2].cnt + tree[cur*2+1].cnt;tree[cur].sum = tree[cur*2].sum + tree[cur*2+1].sum;tree[cur].pfh = tree[cur*2].pfh + tree[cur*2+1].pfh; } void build(int l,int r,int cur) {tree[cur].l=l;tree[cur].r=r;if(l == r) {tree[cur].pfh = tree[cur].sum = tree[cur].cnt = (l==1?n:0);return;}int mid = (l+r)>>1;build(l,mid,cur*2);build(mid+1,r,cur*2+1);pushup(cur); } void update(int cur,ll tar,ll val) {if(tree[cur].l == tree[cur].r) {tree[cur].cnt += val;// 連通塊元素?cái)?shù)量為tar的連通塊的數(shù)量。 tree[cur].sum += val*tar;tree[cur].pfh += val*tar*tar;return;}int mid = (tree[cur].l+tree[cur].r)>>1;if(tar<=mid) update(cur*2,tar,val);else update(cur*2+1,tar,val); pushup(cur); } ll query(int cur,ll k,int sum) {if(tree[cur].l == tree[cur].r) {ll l=1,r=tree[cur].cnt,mid;mid=(l+r)>>1;while(l<r) {mid=(l+r)>>1;if(mid*(mid-1)/2*tree[cur].l*tree[cur].l + mid*tree[cur].l*sum >= k) r=mid;else l = mid+1;}return l;}int mid = (tree[cur].l+tree[cur].r)>>1;ll tmp = (tree[cur*2+1].sum * tree[cur*2+1].sum - tree[cur*2+1].pfh)/2 + tree[cur*2+1].sum * sum ;if(tmp < k)return tree[cur*2+1].cnt + query(cur*2,k-tmp,sum+tree[cur*2+1].sum);else return query(cur*2+1,k,sum); } ll cal(ll k) {if(k<=tot_ret) return cnt;k-=tot_ret;ll tmp = query(1,k,0);return cnt-tmp+1; } int main() {int t;cin>>t;while(t--) {scanf("%d%d",&n,&q);for(int i = 1; i<=n; i++) f[i] = i,num[i] = 1,ret[i] = 0;tot_ret=0,cnt=n;//cnt記錄連通塊數(shù) build(1,n,1);while(q--) {int op;scanf("%d",&op);if(op == 1) {int a,b;scanf("%d%d",&a,&b);a=getf(a),b=getf(b);if(a==b) {ret[a]--;tot_ret--;continue;}if(num[a] > num[b]) swap(a,b);//讓b是邊多的update(1,num[a],-1);update(1,num[b],-1);f[a]=b;//按秩合并到大集合tot_ret += num[a]*num[b]-1;cnt--;ret[b]=ret[a]+ret[b] + num[a]*num[b]-1;num[b]+=num[a];ret[a]=num[a]=0;update(1,num[b],1); }else {ll k;scanf("%lld",&k);ll minn = max(1LL,cnt-k);ll maxx = cal(k);printf("%lld %lld\n",minn,maxx);} }}return 0 ; }

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