HDU3896 Greatest TC(双联通分量+倍增)

Problem Description

TC (Tian Chao) is magical place, as you all know...
The railways and the rail-stations in TC are fragile and always meet with different kinds of problems. In order to reach the destination safely on time, you are asked to develop a system which has two types of main functions as below.
1: A B C D, reporting whether we can get from station A to station B without passing the railway that connects station C and station D.
2: A B C, reporting whether we can get from station A to station B without passing station C.
Please notice that the railways are UNDIRECTED.

 

Input

For each test case, the first line will contain two integers N (2<=N<=100000) and M (1<=M<=500000), namely the number of stations and railways in TC. Then each of the next M lines will have two integers, describing the two stations that a certain railway is connecting. After this, there comes a line containing a single integer Q (Q<=300000), which means the number of queries to make on the system. The next Q lines will be queries. Each query begins with a integer, indicating the type of query, followed by 4 (the first type) or 3 (the second type) integers describing the details of the query as what mentioned above.
The stations are always labeled from 1 to N.

 

Output

For each test case, print "yes" or "no" in separated lines for the queries.

 

Sample Input

13 15

1 2

2 3

3 5

2 4

4 6

2 6

1 4

1 7

7 8

7 9

7 10

8 11

8 12

9 12

12 13

5

1 5 13 1 2

1 6 2 1 4

1 13 6 7 8

2 13 6 7

2 13 6 8

Sample Output

yes

yes

yes

no

yes

 

Source

2011 Multi-University Training Contest 6 - Host by JLU

 

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题解：

无向图，n个点,m条边；

typ==1:

　　a,b,c,d : 判断去掉c和d之间的边之后a,b是否联通；

typ==2:

　　a,b,c:去掉c点之后，a,b是否仍然联通;

思路：先dfs出每个点的dfs序，并处理出每个点的深度以及是否为割点，每条边是否为割边；

对于第一种：（假设c在d的下面）如果a和b一个在c的子树里面，一个不再c的子树里面，并且dc边是桥，则不连通,其他均为联通;

对于第二种：

分成三种情况讨论：

　   1: a,b都在子树c中，如果a,b在c的同一个儿子当中，那么去掉c是联通的；否则让a,b往上跳，变成c的两个儿子，如果lown(a)>=dfn(c) 或 lown(b)>=dfn(c)成立，那么不连通；

　　2：a,b只又一个在子树c中,假设a在子树c中，那么，同样让a往上跳，变成c的儿子,.如果lown[a]>=dfn[c],那么不连通，否则联通;

　　3:a,b都不在子树c中，那么去掉c没有任何影响，所以还是联通(往上跳，可以用倍增法);

 

参考带码：

#include<bits/stdc++.h>
using namespace std;
#define mod 10007
#define pii pair<int,int>
#define pil pair<int,ll>
#define fi first
#define se second
#define mkp make_pair
#define PI acos(-1.0)
typedef long long ll;
const int INF=0x3f3f3f3f;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
    return x*f;
}
inline ll readll()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
    return x*f;
}

const int maxn=1e5+10;
const int maxm=5e5+10;

struct Edge{
    int u,v;
    int nxt;
} edge[maxm<<1];
int n,m,head[maxn],tot,times;
int fa[maxn],dep[maxn],dfn[maxn],out[maxn],lown[maxn];
bool iscut[maxn],isbridge[maxn];
int anc[maxn][20];
inline void AddEdge(int u,int v)
{
    edge[tot].u=u;
    edge[tot].v=v;
    edge[tot].nxt=head[u];
    head[u]=tot++;
}

inline void Init()
{
    tot=times=0;
    memset(head,-1,sizeof(head));
    memset(iscut,false,sizeof(iscut));
    memset(isbridge,false,sizeof(isbridge));
    memset(dep,0,sizeof(dep));
    memset(fa,0,sizeof(fa));
}

inline void dfs(int u)
{
    dfn[u]=lown[u]=++times;
    bool flag=false;
    int child=0;
    for(int e=head[u];~e;e=edge[e].nxt)
    {
        int v=edge[e].v;
        if(v==fa[u]) continue;
        if(!dfn[v])
        {
            fa[v]=u;
            dep[v]=dep[u]+1;
            dfs(v);
            lown[u]=min(lown[u],lown[v]);
            if(lown[v]>=dfn[u])
            {
                iscut[u]=true;
                if(lown[v]>dfn[u]) isbridge[v]=true;
            }
        }
        else lown[u]=min(lown[u],dfn[v]);
    }
    if(u==1 && child==1) iscut[u]=false;
    out[u]=times;
}

inline bool subtree(int x,int y)
{
    return (dfn[x]>=dfn[y]&&out[x]<=out[y])?1:0;
}

inline void preprocess()
{
    memset(anc,0,sizeof(anc));
    for(int i=1;i<=n;++i) anc[i][0]=fa[i];
    for(int j=1;(1<<j)<n;++j)
        for(int i=1;i<=n;++i)
            if(anc[i][j-1]) anc[i][j]=anc[anc[i][j-1]][j-1];
}

inline int upward(int u, int x)
{
    for(int i=0;i<20;i++)
        if((x>>i)&1) u=anc[u][i];
    return u;
}

inline bool Judge(int a,int b,int c)
{
    int in1=subtree(a,c);
    int in2=subtree(b,c);
    if(in1&in2)
    {
        a=upward(a,dep[a]-dep[c]-1);
        b=upward(b,dep[b]-dep[c]-1);
        if(a==b) return true;
        if(lown[a]>=dfn[c]||lown[b]>=dfn[c]) return false;
    }
    if(in1^in2)
    {
        if(!in1) swap(a,b);
        a=upward(a,dep[a]-dep[c]-1);
        if(lown[a]>=dfn[c]) return false;
    }
    return true;
}


int main()
{
    scanf("%d%d",&n,&m);
    Init();
    int u,v;
    for(int i=1;i<=m;++i)
    {
        u=read(),v=read();
        AddEdge(u,v);AddEdge(v,u);
    }
    dfs(1);preprocess();
    int q=read();
    while(q--)
    {
        int typ,a,b,c,d;
        typ=read();a=read();b=read();c=read();
        if(typ==1)
        {
            d=read();
            if(dep[c]<dep[d]) swap(c,d);
            int temp1=subtree(a,c);
            int temp2=subtree(b,c);
            if(isbridge[c]&&(temp1^temp2)) puts("no");
            else puts("yes");
        }
        else
        {
            bool ok=Judge(a,b,c);
            if(ok) puts("yes");else puts("no");
        }
    }
    return 0;
}