【2019浙江省赛 - A】Vertices in the Pocket（权值线段树下二分，图，思维）

题干：

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.

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

Sample Output

3 3
2 3
1 2

解题报告：

以连通块中元素个数为权值，建立线段树，维护三个变量，分别是cnt（这样的连通块的数量），sum（连通块的总的点数），pfh（各连通块之间点数的平方和）（注意不是和的平方）

cnt代表连通块数量，tot_ret代表当前状态下的图，在不增长连通块个数的前提下，可以加的边数。

所以对于最小值很显然。

对于最大值，首先减去块内连的边，然后去线段树查询剩下的边怎么加，首先填连通块元素大小大的，这样也就是类似查询第k大，查到叶子结点，再判断块中元素为tree[cur].l的块我需要多少个返回回去，就行了。

AC代码：

#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;// 连通块元素数量为tar的连通块的数量。 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记录连通块数 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);//让b是边多的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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【2019浙江省赛 - A】Vertices in the Pocket（权值线段树下二分，图，思维）

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