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  • [HNOI 2012] 永无乡

    [题目链接]

             https://www.lydsy.com/JudgeOnline/problem.php?id=2733

    [算法]

            用线段树维护图的联通块 , 并查集维护连通性

            加边时可以直接线段树合并

            时间复杂度 : O(NlogN)

    [代码]

            

    #include<bits/stdc++.h>
    using namespace std;
    #define MAXN 100010 
    #define MAXP 10000010
    typedef long long ll;
    typedef long double ld;
    typedef unsigned long long ull;
    
    int n , m;
    int fa[MAXN] , rt[MAXN] , rk[MAXN] , rev[MAXN];
    
    struct Segment_Tree
    {
            int sz;
            Segment_Tree()
            {
                    sz = 0;
            }
            struct Node
            {
                    int cnt;
                    int lc , rc;
            } a[MAXP];
            inline int merge(int x , int y)
            {
                    if (x == 0 || y == 0) return x + y;
                    a[x].cnt += a[y].cnt;
                    a[x].lc = merge(a[x].lc , a[y].lc);
                    a[x].rc = merge(a[x].rc , a[y].rc);
                    return x;
            }
            inline void update(int x)
            {
                    a[x].cnt = a[a[x].lc].cnt + a[a[x].rc].cnt;
            }
            inline void modify(int &now , int l , int r , int x , int value)
            {
                    if (!now) now = ++sz;
                    if (l == r) 
                    {
                            a[now].cnt = 1;
                            return;
                    }
                    int mid = (l + r) >> 1;
                    if (mid >= x) modify(a[now].lc , l , mid , x , value);
                    else modify(a[now].rc , mid + 1 , r , x , value);
                    update(now);
            }
            inline int query(int &now , int l , int r , int k)
            {
                    if (!now) now = ++sz;
                    if (l == r) return l;
                    int mid = (l + r) >> 1;
                    if (a[a[now].lc].cnt >= k) return query(a[now].lc , l , mid , k);
                    else return query(a[now].rc , mid + 1 , r , k - a[a[now].lc].cnt);
            }
            inline int query(int now , int k)
            {
                    if (a[rt[now]].cnt < k) return -1;
                    else return query(rt[now] , 1 , n , k);
            }
    } SGT;
    
    template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }
    template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); }
    template <typename T> inline void read(T &x)
    {
        T f = 1; x = 0;
        char c = getchar();
        for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
        for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
        x *= f;
    }
    inline int get_root(int x)
    {
            if (fa[x] == x) return x;
            else return fa[x] = get_root(fa[x]);
    }
    
    int main()
    {
            
            scanf("%d%d" , &n , &m);
            for (int i = 1; i <= n; i++) 
            {
                    scanf("%d" , &rk[i]);
                    rev[rk[i]] = i;
            }
            for (int i = 1; i <= n; i++) 
            {
                    fa[i] = i;
                    rt[i] = 0;
            }
            for (int i = 1; i <= n; i++) SGT.modify(rt[i] , 1 , n , rk[i] , 1);
            for (int i = 1; i <= m; i++)
            {
                    int x , y;
                    scanf("%d%d" , &x , &y);
                    if (get_root(x) != get_root(y))
                    {
                            rt[get_root(y)] = SGT.merge(rt[get_root(x)] , rt[get_root(y)]);
                            fa[get_root(x)] = get_root(y);
                    }
            }
            int q;
            scanf("%d" , &q);
            while (q--)
            {
                    char type[5];
                    scanf("%s" , type);
                    if (type[0] == 'Q')
                    {
                            int x , k;
                            scanf("%d%d" , &x , &k);
                            int tmp = SGT.query(get_root(x) , k);
                            if (tmp == -1) printf("-1
    ");    
                            else printf("%d
    " , rev[tmp]);    
                    }    else
                    {
                            int x , y;
                            scanf("%d%d" , &x , &y);
                            if (get_root(x) != get_root(y))
                            {
                                    rt[get_root(y)] = SGT.merge(rt[get_root(x)] , rt[get_root(y)]);
                                    fa[get_root(x)] = get_root(y);
                            }        
                    }    
            }
            
            return 0;
        
    }
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  • 原文地址:https://www.cnblogs.com/evenbao/p/10203407.html
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