hdu 5458 Stability(树链剖分+并查集)

Stability

Time Limit: 3000/2000 MS (Java/Others)    Memory Limit: 65535/102400 K (Java/Others)
Total Submission(s): 1347    Accepted Submission(s): 319

Problem Description

Given an undirected connected graph G with n nodes and m edges, with possibly repeated edges and/or loops. The stability of connectedness between node u and node v is defined by the number of edges in this graph which determines the connectedness between them (once we delete this edge, node u and v would be disconnected).

You need to maintain the graph G, support the deletions of edges (though we guarantee the graph would always be connected), and answer the query of stability for two given nodes.

 

Input

There are multiple test cases(no more than 3 cases), and the first line contains an integer t, meaning the totally number of test cases.

For each test case, the first line contains three integers n, m and q, where 1≤n≤3×104,1≤m≤105 and 1≤q≤105. The nodes in graph G are labelled from 1to n.

Each of the following m lines contains two integers u and v describing an undirected edge between node u and node v.

Following q lines - each line describes an operation or a query in the formats:
⋅ 1 a b: delete one edge between a and b. We guarantee the existence of such edge.
⋅ 2 a b: query the stability between a and b.

 

Output

For each test case, you should print first the identifier of the test case.

Then for each query, print one line containing the stability between corresponding pair of nodes.

 

Sample Input

1
10 12 14
1 2
1 3
2 4
2 5
3 6
4 7
4 8
5 8
6 10
7 9
8 9
8 10
2 7 9
2 7 10
2 10 6
2 10 5
1 10 6
2 10 1
2 10 6
2 3 10
1 8 5
2 5 10
2 4 5
1 7 9
2 7 9
2 10 5

 

Sample Output

Case #1:
0
0
0
0
2
4
3
3
2
3
4

/*
hdu 5458 Stability(树链剖分+并查集)

problem:
给你一个无向图(可能有重边or环). 1.删除a,b之间的边    2.查询a,b的必要边(即删除后a b无法连通)

solve:
最开始想的是建图的时候一直维持一棵树. 如果u,v上面出现环,那么u->v整个链上都可以置为0.因为它们任意两点之间都不可能有必要边
但是后面操作中会删除边,所以不知道树的形状,会导致后面的查询出现问题.

后来考虑倒着操作,先弄出最后操作完的图.本来还担心最后会被切成多个不相连图, 但是
题目：we guarantee the graph would always be connected  保证了操作完后必然是 树或者有环图.
如果是有环图的话可以用过上面的思路转换成树的,最后就成了树链上的操作,所以树链剖分就能做

将所有边记录下来,然后将1操作删除的边除去. 如果最后是有环图(并查集判断)的话,将重复的边记录下,然后建树.
而重复的边会导致 u->v链上面两两没有必要边,于是将置为0.

得出了最后这个树,接着就是倒着操作了
1操作:在u->v上加边 ----> 将u->v整个链上都可以置为0
2操作:查询u->v上的必要边

hhh-2016-08-26 11:06:45
*/
#pragma comment(linker,"/STACK:124000000,124000000")
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <vector>
#include <math.h>
#include <queue>
#include <map>
#include <set>
#define lson  i<<1
#define rson  i<<1|1
#define ll long long
#define clr(a,b) memset(a,b,sizeof(a))
#define scanfi(a) scanf("%d",&a)
#define scanfl(a) scanf("%I64d",&a)
#define key_val ch[ch[root][1]][0]
#define inf 0x3f3f3f3f
using namespace std;
const int maxn = 40010;
const int maxm = 2e5 + 1000;
int head[maxn],tot,pos,son[maxn];
int top[maxn],fp[maxn],fa[maxn],dep[maxn],num[maxn],p[maxn];
int par[maxn];
int n,m,q;
struct Edge
{
    int to,next;
} edge[maxm << 1];
int fin_par(int x)
{
    if(par[x] == -1) return x;
    return par[x] = fin_par(par[x]);
}
void ini()
{
    clr(par,-1);
    tot = 0,pos = 1;
    clr(head,-1),clr(son,-1);
}

void add_edge(int u,int v)
{
    edge[tot].to = v,edge[tot].next = head[u],head[u] = tot++;
}

void dfs1(int u,int pre,int d)
{
//    cout << u << " " <<pre <<" " <<d <<endl;
    dep[u] = d;
    fa[u] = pre,num[u] = 1;
    for(int i = head[u]; ~i; i = edge[i].next)
    {
        int v = edge[i].to;
        if(v != pre)
        {
            dfs1(v,u,d+1);
            num[u] += num[v];
            if(son[u] == -1 || num[v] > num[son[u]])
                son[u] = v;
        }
    }
}

void getpos(int u,int sp)
{
    top[u] = sp;
    p[u] = pos++;
    fp[p[u]] = u;
    if(son[u] == -1)return ;
    getpos(son[u],sp);
    for(int i = head[u]; ~i ; i = edge[i].next)
    {
        int v = edge[i].to;
        if(v != son[u] && v != fa[u])
            getpos(v,v);
    }
}

struct node
{
    int same;
    int val;
    int l,r,mid;
} tree[maxn << 2];

void push_up(int i)
{
    tree[i].val = tree[lson].val + tree[rson].val;
}

void build(int i,int l,int r)
{
    tree[i].same = -1;
    tree[i].l = l,tree[i].r = r;
    tree[i].mid=(l+r) >>1;

    if(l == r)
    {
        if(fp[l] == 1)
            tree[i].val = 0;
        else
            tree[i].val = 1;
//        cout << fp[l]<<":" << tree[i].val <<endl;
        return;
    }
    build(lson,l,tree[i].mid);
    build(rson,tree[i].mid+1,r);
    push_up(i);
}

void push_down(int i)
{
    if(tree[i].same != -1)
    {
        tree[lson].same = tree[i].same;
        tree[rson].same = tree[i].same;
        tree[lson].val = tree[rson].val = 0;
        tree[i].same = -1;
    }
}

void update_area(int i,int l,int r)
{
    if(tree[i].l >= l && tree[i].r <= r)
    {
        tree[i].val= 0;
        tree[i].same=  0;
        return ;
    }
    push_down(i);
    if(l <= tree[i].mid)
        update_area(lson,l,r);
    if(r > tree[i].mid)
        update_area(rson,l,r);
    push_up(i);
    return ;
}

int query(int i,int l,int r)
{
    if(tree[i].l >= l && tree[i].r <= r)
        return tree[i].val;
    int tcnt = 0;
    push_down(i);
    if(l <= tree[i].mid)
        tcnt += query(lson,l,r);
    if(r > tree[i].mid)
        tcnt += query(rson,l,r);
    push_up(i);
    return tcnt;
}

void make_update(int u,int v)
{
    int f1 = top[u],f2 = top[v];
    while(f1 != f2)
    {
        if(dep[f1] < dep[f2])swap(f1,f2),swap(u,v);
        update_area(1,p[f1],p[u]);
        u = fa[f1] , f1 =top[u];
    }
    if(u == v)
        return ;
    if(dep[u] > dep[v])  swap(u,v);
    update_area(1,p[son[u]],p[v]);
}

int make_query(int u,int v)
{
    int ans = 0;
    int f1= top[u],f2 = top[v];
    while(f1 != f2)
    {
        if(dep[f1] < dep[f2])
        {
            swap(f1,f2),swap(u,v);
        }
        ans += query(1,p[f1],p[u]);
//        cout << p[f1] <<" " <<p[u] <<" "<<ans << endl;
        u = fa[f1] , f1 = top[u];
    }
    if(u == v)
        return ans;
    if(dep[u] > dep[v]) swap(u,v);
    ans += query(1,p[son[u]],p[v]);
//    cout << p[u]+1 <<" " <<p[v] <<" "<<ans << endl;
    return ans ;
}

int l[maxm],r[maxm],op[maxm];
int ans[maxm];
struct Point
{
    int l,r;
    Point() {}
    Point(int x,int y):l(x),r(y) {}
    bool operator <(const Point &a)const
    {
        if(l != a.l)
            return l < a.l;
        else
            return r < a.r;
    }
};
multiset<Point> vec;
multiset<Point> another;
multiset<Point>::iterator it;
int main()
{
//    freopen("in.txt","r",stdin);
    int T,u,v,cas = 1;
    scanfi(T);
    while(T--)
    {
        ini();
        vec.clear(),another.clear();
        scanfi(n),scanfi(m),scanfi(q);
        for(int i = 1; i <= m; i++)
        {
            scanfi(u),scanfi(v);
            if(u > v) swap(u,v);
            vec.insert(Point(u,v));
        }
        for(int i = 1; i <= q; i++)
        {
            scanf("%d%d%d",&op[i],&l[i],&r[i]);
            if(op[i] == 1)
            {
                if(l[i] > r[i]) swap(l[i],r[i]);
                vec.erase(vec.lower_bound(Point(l[i],r[i])));
            }
        }
//        cout << vec.size() <<endl;
        for( it = vec.begin(); it != vec.end(); it++)
        {
            Point t = *it;
            u = t.l,v = t.r;
            int tu = fin_par(u),tv = fin_par(v);
//            cout << u <<" " << v << " " << tu << " "  <<tv <<endl;
            if(tu == tv)
            {
                another.insert(Point(u,v));
                continue;
            }
            par[tu] = tv;
            add_edge(u,v);
            add_edge(v,u);
        }
//        cout << another.size() <<endl;
        dfs1(1,0,0);
        getpos(1,1);
        build(1,1,pos-1);
        printf("Case #%d:\n",cas++);
        for( it = another.begin(); it != another.end(); it++)
        {
            Point t = *it;
            u = t.l,v = t.r;
            make_update(u,v);
        }
        int ob = 0;
        for(int i = q; i > 0; i--)
        {
            u = l[i],v = r[i];
//            cout << op[i]<<" "<< u << " " << v <<endl;

            if(op[i] == 2)
            {
                ans[ob++] = make_query(u,v);
            }
            else
            {
                make_update(u,v);
            }
        }
        for(int i = ob-1;i >=0;i--)
        {
            printf("%d\n",ans[i]);
        }
    }
    return 0;
}