BZOJ3674 可持久化并査集

@(BZOJ)[可持久化并査集]

Description

n个集合 m个操作
操作：
1 a b 合并a,b所在集合
2 k 回到第k次操作之后的状态(查询算作操作)
3 a b 询问a,b是否属于同一集合，是则输出1否则输出0
请注意本题采用强制在线,所给的a,b,k均经过加密,加密方法为x = x xor lastans,lastans的初始值为0
0<n,m<=2*10^5

Sample Input

5 6
1 1 2
3 1 2
2 1
3 0 3
2 1
3 1 2

Sample Output

1
0
1

Solution

模板题.
BZOJ上的数据傻到不行.
还是去xsy上提交吧

#include <cstdio>
#include <cctype>
#include <cstring>
#include <algorithm>

const int N = 1 << 18, M = 1 << 18;
int n;

namespace Zeonfai
{
    inline int getInt()
    {
        int a = 0, sgn = 1;
        char c;

        while(! isdigit(c = getchar()))
            sgn *= c == '-' ? -1 : 1;

        while(isdigit(c))
            a = a * 10 + c - '0', c = getchar();

        return a * sgn;
    }
}

struct persistentDisjointSet
{
    int rt[M], tp;

    struct data
    {
        int pre, dep;
    };

    struct node
    {
        int suc[2];
        data infr;
    }nd[N * 18 + M * 18];

    inline int newNode(int pre, int dep)
    {
        nd[tp].suc[0] = nd[tp].suc[1] = -1;
        nd[tp].infr.pre = pre, nd[tp].infr.dep = dep;
        return tp ++;
    }

    int build(int L, int R)
    {
        int u = newNode(L == R ? L : -1, 1);
        
        if(L == R)
            return u;
        
        int mid = L + R >> 1;
        nd[u].suc[0] = build(L, mid), nd[u].suc[1] = build(mid + 1, R);
        return u;
    }

    inline void initialize()
    {
        memset(rt, -1, sizeof(rt));
        tp = 0;
        rt[0] = build(1, n);
    }

    int find(int L, int R, int u, int a)
    {
        if(L == R)
            return u;

        int mid = L + R >> 1;
        
        if(a <= mid)
            return find(L, mid, nd[u].suc[0], a);
        else
            return find(mid + 1, R, nd[u].suc[1], a);
    }

    inline int access(int tm, int a)
    {
        while(int u = find(1, n, rt[tm], a))
        {
            if(nd[u].infr.pre == a)
                return a;
            
            a = nd[u].infr.pre;
        }
    }

    int newSegmentTree(int _u, int L, int R, int a, int b)
    {
        int u = tp ++;
        nd[u] = nd[_u];
        
        if(L == R)
            return u;
        
        int mid = L + R >> 1;

        if((a >= L && a <= mid) || (b >= L && b <= mid))
            nd[u].suc[0] = newSegmentTree(nd[_u].suc[0], L, mid, a, b);

        if((a > mid && a <= R) || (b > mid && b <= R))
            nd[u].suc[1] = newSegmentTree(nd[_u].suc[1], mid + 1, R, a, b);
        
        return u;
    }

    inline void merge(int tm, int a, int b)
    {
        int rtA = access(tm - 1, a), rtB = access(tm - 1, b);

        if(rtA == rtB)
        {
            rt[tm] = rt[tm - 1];
            return;
        }

        rt[tm] = newSegmentTree(rt[tm - 1], 1, n, rtA, rtB);
        int u = find(1, n, rt[tm], rtA), v = find(1, n, rt[tm], rtB);

        if(nd[u].infr.dep == nd[v].infr.dep)
        {
            ++ nd[u].infr.dep;
            nd[v].infr.pre = rtA;
            return;
        }

        if(nd[u].infr.dep > nd[v].infr.dep)
            nd[v].infr.pre = rtA;
        else
            nd[u].infr.pre = rtB;
    }
    
    inline void accessHistory(int tm, int _tm)
    {
        rt[tm] = rt[_tm];
    }

    inline int query(int tm, int a, int b)
    {
        rt[tm] = rt[tm - 1];
        int preA = access(tm, a), preB = access(tm, b);

        if(preA == preB)
            return 1;
        else
            return 0;
    }
}st;

int main()
{
    #ifndef ONLINE_JUDGE
    freopen("BZOJ3674.in", "r", stdin);
    freopen("BZOJ3674.out", "w", stdout);
    #endif
    
    using namespace Zeonfai;

    n = getInt();
    int m = getInt(), lstAns = 0, tm = 0;
    st.initialize();

    while(m --)
    {
        int opt = getInt();

        if(opt == 1)
        {
            int a = getInt() ^ lstAns, b = getInt() ^lstAns;
            st.merge(++ tm, a, b);
        }
        else if(opt == 2)
        {
            int a = getInt() ^ lstAns;
            st.accessHistory(++ tm, a);
        }
        else if(opt == 3)
        {
            int a = getInt() ^ lstAns, b = getInt() ^ lstAns;
            printf("%d\n", lstAns = st.query(++ tm, a, b));
        }
    }
}

转载于:https://www.cnblogs.com/ZeonfaiHo/p/6681655.html