Codeforces Manthan, Codefest 18 (rated, Div. 1 + Div. 2) D，E

D. Valid BFS?

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

The BFS algorithm is defined as follows.

1. Consider an undirected graph with vertices numbered from $\mathrm{11 to nn.\; Initialize qq as\; a\; new queue containing\; only\; vertex 11,\; mark\; the\; vertex 11 as\; used.}$
2. Extract a vertex $\mathrm{vv from\; the\; head\; of\; the\; queue qq.}$
3. Print the index of vertex $\mathrm{vv.}$
4. Iterate in arbitrary order through all such vertices $\mathrm{uu that uu is\; a\; neighbor\; of vv and\; is\; not\; marked\; yet\; as\; used.\; Mark\; the\; vertex uu as\; used\; and\; insert\; it\; into\; the\; tail\; of\; the\; queue qq.}$
5. If the queue is not empty, continue from step 2.
6. Otherwise finish.

Since the order of choosing neighbors of each vertex can vary, it turns out that there may be multiple sequences which BFS can print.

In this problem you need to check whether a given sequence corresponds to some valid BFS traversal of the given tree starting from vertex $\mathrm{11.\; The tree is\; an\; undirected\; graph,\; such\; that\; there\; is\; exactly\; one\; simple\; path\; between\; any\; two\; vertices.}$

Input

The first line contains a single integer $\mathrm{nn (1\le n\le 2\cdot 1051\le n\le 2\cdot 105)\; which\; denotes\; the\; number\; of\; nodes\; in\; the\; tree.}$

The following $\mathrm{n-1n-1 lines\; describe\; the\; edges\; of\; the\; tree.\; Each\; of\; them\; contains\; two\; integers xx and yy (1\le x,y\le n1\le x,y\le n) —\; the\; endpoints\; of\; the\; corresponding\; edge\; of\; the\; tree.\; It\; is\; guaranteed\; that\; the\; given\; graph\; is\; a\; tree.}$

The last line contains $\mathrm{nn distinct\; integers a1,a2,\dots ,ana1,a2,\dots ,an (1\le ai\le n1\le ai\le n) —\; the\; sequence\; to\; check.}$

Output

Print "Yes" (quotes for clarity) if the sequence corresponds to some valid BFS traversal of the given tree and "No" (quotes for clarity) otherwise.

You can print each letter in any case (upper or lower).

Examples

input

Copy

4
1 2
1 3
2 4
1 2 3 4

output

Yes

input

4
1 2
1 3
2 4
1 2 4 3

output

No

Note

Both sample tests have the same tree in them.

In this tree, there are two valid BFS orderings:

• $\mathrm{1,2,3,41,2,3,4,}$
• $\mathrm{1,3,2,41,3,2,4.}$

The ordering $\mathrm{1,2,4,31,2,4,3 doesn\text{'}t\; correspond\; to\; any\; valid BFS order.}$

题意  给定一棵树，在给定一个序列，问是不是以1为根的BFS序。

解析  BFS序的特点就是 安层次的所以  i 的儿子是连续的 直接暴力判断当前的段是不是都是 i 的儿子。

#include <bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define all(a) (a).begin(), (a).end()
#define fillchar(a, x) memset(a, x, sizeof(a))
#define huan printf("
");
#define debug(a,b) cout<<a<<" "<<b<<" ";
using namespace std;
const int maxn=2e5+10;
typedef long long ll;
vector<int> g[maxn];
map<pair<int,int>,int> ma;
int a[maxn];
int b[maxn];
int main()
{
    int n;
    cin>>n;
    for(int i=0;i<n-1;i++)
    {
        int u,v;
        cin>>u>>v;
        g[u].pb(v);
        g[v].pb(u);
        ma[mp(u,v)]=1;
        ma[mp(v,u)]=1;
    }
    for(int i=1;i<=n;i++)
        cin>>a[i];
    if(a[1]!=1)
    {
        cout<<"No"<<endl;
        return 0;
    }
    int flag=1,before=1;
    for(int i=1;i<n;i++)
    {
        int num=0;
        for(int j=0;j<g[a[i]].size();j++)
            if(b[g[a[i]][j]]==0)num++;
        //cout<<i<<" "<<num<<endl;
        for(int j=0;j<num;j++)
        {
            if(before+j+1>n)
                break;
           // cout<<a[i+j+1]<<" j"<<j<<endl;
            if(ma[mp(a[i],a[before+j+1])]!=1)
            {
                flag=0;
                break;
            }
            //else
              //  b[a[i+j+1]]=1;
        }
        before=before+num;
        //cout<<a[i]<<" "<<flag<<endl;
        b[a[i]]=1;
        if(flag==0)
            break;
    }
    if(flag)
        cout<<"Yes"<<endl;
    else
        cout<<"No"<<endl;
}

E. Trips

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

There are $\mathrm{nn persons\; who\; initially\; don\text{'}t\; know\; each\; other.\; On\; each\; morning,\; two\; of\; them,\; who\; were\; not\; friends\; before,\; become\; friends.}$

We want to plan a trip for every evening of $\mathrm{mm days.\; On\; each\; trip,\; you\; have\; to\; select\; a\; group\; of\; people\; that\; will\; go\; on\; the\; trip.\; For\; every\; person,\; one\; of\; the\; following\; should\; hold:}$

• Either this person does not go on the trip,
• Or at least $\mathrm{kk of\; his\; friends\; also\; go\; on\; the\; trip.}$

Note that the friendship is not transitive. That is, if $\mathrm{aa and bb are\; friends\; and bb and cc are\; friends,\; it\; does\; not\; necessarily\; imply\; that aa and cc are\; friends.}$

For each day, find the maximum number of people that can go on the trip on that day.

Input

The first line contains three integers $\mathrm{nn, mm,\; and kk (2\le n\le 2\cdot 105,1\le m\le 2\cdot 1052\le n\le 2\cdot 105,1\le m\le 2\cdot 105, 1\le k<n1\le k<n) —\; the\; number\; of\; people,\; the\; number\; of\; days\; and\; the\; number\; of\; friends\; each\; person\; on\; the\; trip\; should\; have\; in\; the\; group.}$

The $\mathrm{ii-th\; (1\le i\le m1\le i\le m)\; of\; the\; next mm lines\; contains\; two\; integers xx and yy (1\le x,y\le n1\le x,y\le n, x\ne yx\ne y),\; meaning\; that\; persons xx and yy become\; friends\; on\; the\; morning\; of\; day ii.\; It\; is\; guaranteed\; that xx and yy were\; not\; friends\; before.}$

Output

Print exactly $\mathrm{mm lines,\; where\; the ii-th\; of\; them\; (1\le i\le m1\le i\le m)\; contains\; the\; maximum\; number\; of\; people\; that\; can\; go\; on\; the\; trip\; on\; the\; evening\; of\; the\; day ii.}$

Examples

input

Copy

4 4 2
2 3
1 2
1 3
1 4

output

Copy

0
0
3
3

input

Copy

5 8 2
2 1
4 2
5 4
5 2
4 3
5 1
4 1
3 2

output

Copy

0
0
0
3
3
4
4
5

input

Copy

5 7 2
1 5
3 2
2 5
3 4
1 2
5 3
1 3

output

Copy

0
0
0
0
3
4
4

Note

In the first example,

• $\mathrm{1,2,31,2,3 can\; go\; on\; day 33 and 44.}$

In the second example,

• $\mathrm{2,4,52,4,5 can\; go\; on\; day 44 and 55.}$
• $\mathrm{1,2,4,51,2,4,5 can\; go\; on\; day 66 and 77.}$
• $\mathrm{1,2,3,4,51,2,3,4,5 can\; go\; on\; day 88.}$

In the third example,

• $\mathrm{1,2,51,2,5 can\; go\; on\; day 55.}$
• $\mathrm{1,2,3,51,2,3,5 can\; go\; on\; day 66 and 77.}$

$\mathrm{解析\; 我们离线处理这个问题，先把每个点的入度和编号用pair保存起来\; 扔到set里面\; ，每次把度数小于k的点删掉，与它相邻的点\; j\; 度数减1\; 。把原来在set里的pair<du[j],j>删掉，再插入新的点}$

$\mathrm{直到\; set里的点的度数都大于k。set的size就是当前的答案。删掉当前的边，重复上面的操作。}$

#include <bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define all(a) (a).begin(), (a).end()
#define fillchar(a, x) memset(a, x, sizeof(a))
#define huan printf("
");
#define debug(a,b) cout<<a<<" "<<b<<" ";
using namespace std;
const int maxn=2e5+10;
typedef long long ll;
typedef pair<int,int> pii;
int du[maxn];
set<int> g[maxn];
int b1[maxn],b2[maxn],ans[maxn],vis[maxn];
int main()
{
    int n,m,k;
    scanf("%d%d%d",&n,&m,&k);
    set<pii> s;
    for(int i=0;i<m;i++)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        b1[i]=x,b2[i]=y;
        du[x]++,du[y]++;
        g[x].insert(y),g[y].insert(x);
    }
    for(int i=1;i<=n;i++)
        s.insert(mp(du[i],i));
    for(int i=m-1;i>=0;i--)
    {
        while(s.size()>=1)
        {
            auto itt=(*s.begin());
            if(itt.fi>=k)break;
            for(auto it=g[itt.se].begin();it!=g[itt.se].end();it++)
            {
                int u=*it;
                if(vis[u]==1)continue;
                s.erase(mp(du[u],u));
                du[u]--;
                s.insert(mp(du[u],u));
                g[u].erase(itt.se);
            }
            s.erase(itt);
            vis[itt.se]=1;
        }
        ans[i]=s.size();
        if(vis[b1[i]]==0&&vis[b2[i]]==0)
        {
            s.erase(mp(du[b1[i]],b1[i]));
            du[b1[i]]--;
            s.insert(mp(du[b1[i]],b1[i]));
            g[b1[i]].erase(b2[i]);
            s.erase(mp(du[b2[i]],b2[i]));
            du[b2[i]]--;
            s.insert(mp(du[b2[i]],b2[i]));
            g[b2[i]].erase(b1[i]);
        }
    }
    for(int i=0;i<m;i++)
        printf("%d
",ans[i]);
}