HDU 5333 Undirected Graph【LCT+BIT】

àì夳堔傛蜴生んèń 2022-08-04 05:26 134阅读 0赞

LCT：  
每次操作相当于只把区间\[L,R\]之间的边连起来，求联通分量的个数。  
思路：  
把操作排序后，对于区间\[L,R\]的操作，先把R所有v<R的边(R→v)加入集合， 用动态树+并查集维护\[1,R\]的联通分量的个数cnt  
那么，答案==\[L,R\]区间的联通分量的个数\+N−R\+L−1；  
求区间\[L,R\]的联通分量的个数，可以考虑把所有\[1,L−1\]的边删掉后新形成的联通分量。  
假设当前的连通分量x为一棵树，那么删掉一条边肯定会形成一个新的连通分量，所以删掉所有\[1,L\]的边形成的连通分量的个数即  
这棵树上所有(u→v且v<L)的边的数量。  
如果当前联通分量是一棵树+一条边的话， 那么，如果树中有|x|−1条包含在\[L,R\]内的边，则不会形成新的连通分量，  
考虑到删掉的边的单调性，即如果对于一棵树加入一条新的边（u→v），那么这条边可以代替所有（u′→v′且v′<v）的边，  
也就是说v’这条边就算在之后的操作中会被删掉也不会对答案产生影响，因此删掉这条边，并且将新边加入此树。  
故区间\[L,R\]的连通分量的个数即cnt\+当前森林中(u→v,v<L)的边的数量；  
前者可以用LCT+并查集维护，后者可以用树状数组维护。  
对于当前新加入的边(u→v)如果u和v不在同一个连通分量，则用并查集合并他们并且将他们的LCT合并。  
否则的话， 将u所在的LCT中u→v这条链的边里面两点最小值最小的那一条边删除，并且在树状数组中将对应点也删除。

[(代码来源)][Link 1]

#pragma comment(linker, "/STACK:1024000000")
    #include <stdio.h>
    #include <string.h>
    #include <map>
    #include <algorithm>
    using namespace std ;
    
    typedef long long LL ;
    
    #define clr( a , x ) memset ( a , x , sizeof a )
    #define ls ( o << 1 )
    #define rs ( o << 1 | 1 )
    #define lson ls , l , m
    #define rson rs , m + 1 , r
    #define root 1 , 1 , n
    #define mid ( ( l + r ) >> 1 )
    
    const int MAXN = 100005 ;
    const int MAXE = 200005 ;
    const int INF = 0x3f3f3f3f ;
    
    struct Edge {
        int v , idx , n ;
        Edge () {}
        Edge ( int v , int idx , int n ) : v ( v ) , idx ( idx ) , n ( n ) {}
    } ;
    
    struct Node* null ;
    
    struct Node {
        Node* c[2] ;
        Node* f ;
        bool flip ;
        int minv , val ;
        int eidx , idx ;
        void newnode ( int v , int i ) {
            c[0] = c[1] = f = null ;
            minv = val = v ;
            eidx = idx = i ;
            flip = 0 ;
        }
        void rev () {
            if ( this == null ) return ;
            swap ( c[0] , c[1] ) ;
            flip ^= 1 ;
        }
        void up () {
            if ( this == null ) return ;
            if ( val <= c[0]->minv && val <= c[1]->minv ) {
                minv = val ;
                eidx = idx ;
            } else if ( c[0]->minv <= c[1]->minv && c[0]->minv <= val ) {
                minv = c[0]->minv ;
                eidx = c[0]->eidx ;
            } else {
                minv = c[1]->minv ;
                eidx = c[1]->eidx ;
            }
        }
        void down () {
            if ( this == null ) return ;
            if ( flip ) {
                c[0]->rev () ;
                c[1]->rev () ;
                flip = 0 ;
            }
        }
        bool is_root () {
            return f == null || f->c[0] != this && f->c[1] != this ;
        }
        void sign_down () {
            if ( !is_root () ) f->sign_down () ;
            down () ;
        }
        void setc ( Node* o , int d ) {
            c[d] = o ;
            o->f = this ;
        }
        void rot ( int d ) {
            Node* p = f ;
            Node* g = f->f ;
            p->setc ( c[d] , !d ) ;
            if ( !p->is_root () ) g->setc ( this , f == g->c[1] ) ;
            else f = g ;
            setc ( p , d ) ;
            p->up () ;
        }
        void splay () {
            sign_down () ;
            while ( !is_root () ) {
                if ( f->is_root () ) rot ( this == f->c[0] ) ;
                else {
                    if ( f == f->f->c[0] ) {
                        if ( this == f->c[0] ) f->rot ( 1 ) , rot ( 1 ) ;
                        else rot ( 0 ) , rot ( 1 ) ;
                    } else {
                        if ( this == f->c[1] ) f->rot ( 0 ) , rot ( 0 ) ;
                        else rot ( 1 ) , rot ( 0 ) ;
                    }
                }
            }
            up () ;
        }
        void access () {
            Node* o = this ;
            for ( Node* x = null ; o != null ; x = o , o = o->f ) {
                o->splay () ;
                o->setc ( x , 1 ) ;
                o->up () ;
            }
            splay () ;
        }
        void make_root () {
            access () ;
            rev () ;
        }
        void link ( Node* o ) {
            make_root () ;
            f = o ;
        }
        void cut () {
            access () ;
            c[0] = c[0]->f = null ;
            up () ;
        }
        void cut ( Node* o ) {
            make_root () ;
            o->cut () ;
        }
        int get_min ( Node* o ) {
            make_root () ;
            o->access () ;
            return o->eidx ;
        } 
    } ;
    
    Node pool[MAXN + MAXE] ;
    Node* cur ;
    Node* node[MAXN] ;
    Node* edge[MAXE] ;
    
    Edge E[MAXE + MAXN] ;
    int H[MAXN] , Q[MAXN] , cntE ;
    
    int U[MAXE] , V[MAXE] ;
    int ans[MAXN] ;
    int p[MAXN] ;
    int c[MAXN] ;
    int n , m , q ;
    
    int find ( int x ) {
        return p[x] == x ? x : ( p[x] = find ( p[x] ) ) ;
    }
    
    void init ( int n ) {
        cntE = 0 ;
        cur = pool ;
        cur->newnode ( INF , -1 ) ;
        null = cur ++ ;
        for ( int i = 1 ; i <= n ; ++ i ) {
            p[i] = i ;
            H[i] = -1 ;
            Q[i] = -1 ;
            c[i] = 0 ;
            cur->newnode ( INF , -1 ) ;
            node[i] = cur ++ ;
        }
    }
    
    void addedge ( int u , int v , int idx , int H[] ) {
        E[cntE] = Edge ( v , idx , H[u] ) ;
        H[u] = cntE ++ ;
    }
    
    void add ( int x , int v ) {
        for ( ; x <= n ; x += x & -x ) c[x] += v ;
    }
    
    int sum ( int x , int ans = 0 ) {
        for ( ; x > 0 ; x -= x & -x ) ans += c[x] ;
        return ans ;
    }
    
    void scanf ( int& x , char c = 0 ) {
        while ( ( c = getchar () ) < '0' ) ;
        x = c - '0' ;
        while ( ( c = getchar () ) >= '0' ) x = x * 10 + c - '0' ;
    }
    
    void solve () {
        int cnt = 0 ;
        int u , v ;
        init ( n ) ;
        for ( int i = 1 ; i <= n ; ++ i ) add ( i , -1 ) ;
        for ( int i = 0 ; i < m ; ++ i ) {
            scanf ( "%d%d" , &u , &v ) ;
            if ( u == v ) continue ;
            if ( u < v ) swap ( u , v ) ;
            addedge ( u , v , i , H ) ;
            U[i] = u ;
            V[i] = v ;
            cur->newnode ( v , i ) ;
            edge[i] = cur ++ ;
        }
        for ( int i = 0 ; i < q ; ++ i ) {
            scanf ( "%d%d" , &u , &v ) ;
            addedge ( v , u , i , Q ) ;
        }
        for ( int i = 1 ; i <= n ; ++ i ) {
            ++ cnt ;
            for ( int j = H[i] ; ~j ; j = E[j].n ) {
                int v = E[j].v , idx = E[j].idx ;
                if ( v == i ) continue ;
                int x = find ( i ) ;
                int y = find ( v ) ;
                if ( x != y ) {
                    -- cnt ;
                    p[x] = y ;
                    edge[idx]->link ( node[i] ) ;
                    edge[idx]->link ( node[v] ) ;
                    add ( v , 1 ) ;
                } else {
                    int eidx = node[i]->get_min ( node[v] ) ;
                    if ( V[eidx] >= v ) continue ;
                    edge[eidx]->cut ( node[U[eidx]] ) ;
                    edge[eidx]->cut ( node[V[eidx]] ) ;
                    add ( V[eidx] , -1 ) ;
                    edge[idx]->link ( node[i] ) ;
                    edge[idx]->link ( node[v] ) ;
                    add ( v , 1 ) ;
                }
            }
            for ( int j = Q[i] ; ~j ; j = E[j].n ) {
                int v = E[j].v , idx = E[j].idx ;
                ans[idx] = cnt + sum ( v - 1 ) + n - i + v - 1 ;
            }
        }
        for ( int i = 0 ; i < q ; ++ i ) {
            printf ( "%d\n" , ans[i] ) ;
        }
    }
    
    int main () {
        while ( ~scanf ( "%d%d%d" , &n , &m , &q ) ) solve () ;
        return 0 ;
    }

不同的姿势吧，还是推荐上面那种封装好的，比较方便

// whn6325689
    // Mr.Phoebe
    // http://blog.csdn.net/u013007900
    #include <algorithm>
    #include <iostream>
    #include <iomanip>
    #include <cstring>
    #include <climits>
    #include <complex>
    #include <fstream>
    #include <cassert>
    #include <cstdio>
    #include <bitset>
    #include <vector>
    #include <deque>
    #include <queue>
    #include <stack>
    #include <ctime>
    #include <set>
    #include <map>
    #include <cmath>
    #include <functional>
    #include <numeric>
    #pragma comment(linker, "/STACK:1024000000,1024000000")
    
    using namespace std;
    
    #define eps 1e-9
    #define PI acos(-1.0)
    #define INF 0x3f3f3f3f
    #define LLINF 1LL<<62
    #define speed std::ios::sync_with_stdio(false);
    
    typedef long long ll;
    typedef unsigned long long ull;
    typedef long double ld;
    typedef pair<ll, ll> pll;
    typedef complex<ld> point;
    typedef pair<int, int> pii;
    typedef pair<pii, int> piii;
    typedef vector<int> vi;
    
    #define CLR(x,y) memset(x,y,sizeof(x))
    #define CPY(x,y) memcpy(x,y,sizeof(x))
    #define clr(a,x,size) memset(a,x,sizeof(a[0])*(size))
    #define cpy(a,x,size) memcpy(a,x,sizeof(a[0])*(size))
    
    #define mp(x,y) make_pair(x,y)
    #define pb(x) push_back(x)
    #define lowbit(x) (x&(-x))
    
    #define MID(x,y) (x+((y-x)>>1))
    #define ls (idx<<1)
    #define rs (idx<<1|1)
    #define lson ls,l,mid
    #define rson rs,mid+1,r
    #define root 1,1,n
    
    template<class T>
    inline bool read(T &n)
    {
        T x = 0, tmp = 1;
        char c = getchar();
        while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();
        if(c == EOF) return false;
        if(c == '-') c = getchar(), tmp = -1;
        while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();
        n = x*tmp;
        return true;
    }
    template <class T>
    inline void write(T n)
    {
        if(n < 0)
        {
            putchar('-');
            n = -n;
        }
        int len = 0,data[20];
        while(n)
        {
            data[len++] = n%10;
            n /= 10;
        }
        if(!len) data[len++] = 0;
        while(len--) putchar(data[len]+48);
    }
    //-----------------------------------
    
    const int maxn = 300010;
    
    struct Query
    {
        int a, b, i;
        bool operator < (const Query &rhs) const
        {
            return b < rhs.b;
        }
        void input(int ii = 0)
        {
            scanf("%d%d", &a, &b);
            if(a > b) swap(a, b);
            i = ii;
        }
    } que[maxn], e[maxn];
    
    int n, m, q, ans[maxn], block_tot;
    int ch[maxn][2], fa[maxn], rt[maxn], miv[maxn];
    int val[maxn], sum[maxn], rev[maxn];
    
    void ins(int x, int c)
    {
        for(int i = x; i <= n; i += i&(-i))
            sum[i] += c;
    }
    
    int getsum(int x)
    {
        int ans = 0;
        for(int i = x; i; i -= i&(-i))
            ans += sum[i];
        return ans;
    }
    
    void update_rev(int r)
    {
        if(!r) return;
        swap(ch[r][0], ch[r][1]);
        rev[r] ^= 1;
    }
    
    void push_up(int x)
    {
        int lc = ch[x][0], rc = ch[x][1];
        miv[x] = x;
        if(val[miv[lc]] < val[miv[x]]) miv[x] = miv[lc];
        if(val[miv[rc]] < val[miv[x]]) miv[x] = miv[rc];
    }
    
    void push_down(int r)
    {
        if(rev[r])
        {
            update_rev(ch[r][0]);
            update_rev(ch[r][1]);
            rev[r] = 0;
        }
    }
    
    bool judge(int u, int v)
    {
        while(fa[u]) u = fa[u];
        while(fa[v]) v = fa[v];
        return u == v;
    }
    
    void rotate(int x)
    {
        int y = fa[x], kind = ch[y][1]==x;
        ch[y][kind] = ch[x][!kind];
        fa[ch[y][kind]] = y;
        fa[x] = fa[y];
        fa[y] = x;
        ch[x][!kind] = y;
        if(rt[y])
            rt[y] = 0, rt[x] = 1;
        else
            ch[fa[x]][ch[fa[x]][1]==y] = x;
        push_up(y);
    }
    
    void P(int r)
    {
        if(!rt[r]) P(fa[r]);
        push_down(r);
    }
    
    void splay(int x)
    {
        P(x);
        while(!rt[x])
        {
            int f = fa[x], ff = fa[f];
            if(rt[f])
                rotate(x);
            else if((ch[ff][1]==f)==(ch[f][1]==x))
                rotate(f), rotate(x);
            else
                rotate(x), rotate(x);
        }
        push_up(x);
    }
    
    void access(int x)
    {
        // return;
        for(int t = 0; x; t = x, x = fa[x])
        {
            splay(x);
            rt[ch[x][1]] = 1;
            rt[ch[x][1]=t] = 0;
            push_up(x);
        }
    }
    
    void makeroot(int x)
    {
        access(x);
        splay(x);
        update_rev(x);
    }
    
    void cut(int x, int y)
    {
        makeroot(x);
        splay(y);
        fa[ch[y][0]] = fa[y];
        fa[y] = 0;
        rt[ch[y][0]] = 1;
        ch[y][0] = 0;
        push_up(y);
    }
    
    void link(int x, int y)
    {
        makeroot(x);
        fa[x] = y;
    }
    
    int query(int x, int y)
    {
        makeroot(x);
        access(y);
        splay(y);
        return miv[y];
    }
    
    void gao(int u, int v, int i)
    {
        // printf("%d %d :: %d %d", u, v, fa[u], fa[v]); cout << endl;
        if(judge(u, v))
        {
            int t = query(u, v);
            if(u == v || u <= val[t]) return;
            block_tot++;
            cut(t, e[t-n-1].a);
            cut(t, e[t-n-1].b);
            ins(val[t], -1);
        }
        block_tot--;
        val[n+i] = u;
        miv[n+i] = n+i;
        link(u, n+i);
        link(v, n+i);
        ins(u, 1);
        // for(int j = 1; j <= n+m; j++)
        // printf("%d ", fa[j]);
        // cout << "return" << i << ' ' << u << ' ' << v << ' ' << block_tot << endl;
    }
    
    int main()
    {
        while(~scanf("%d%d%d", &n, &m, &q))
        {
            for(int i = 0; i <= n; i++)
                val[i] = INF, sum[i] = 0;
            for(int i = 0; i <= n+m; i++)
                rev[i] = ch[i][0] = ch[i][1] = fa[i] = 0, rt[i] = 1;
    
            for(int i = 0; i < m; i++)
                e[i].input();
            sort(e, e+m);
            for(int i = 0; i < q; i++)
                que[i].input(i);
            sort(que, que+q);
    
            block_tot = n;
            for(int i = 0, j = 0; i < q; i++)
            {
                int l = que[i].a, r = que[i].b;
                while(j < m && e[j].b <= r)
                {
                    gao(e[j].a, e[j].b, j+1);
                    j++;
                    // printf("%d %d___\n", j, m);
                }
                ans[que[i].i] = block_tot + getsum(l-1);
                // printf("%d %d__%d\n", l-1, block_tot, que[i].i);
            }
            for(int i = 0; i < q; i++)
                printf("%d\n", ans[i]);
        }
        return 0;
    }

[Link 1]: http://blog.csdn.net/u013368721/article/details/47168603