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  • 洛谷P1501 [国家集训队]Tree II(LCT)

    题目描述

    一棵n个点的树,每个点的初始权值为1。对于这棵树有q个操作,每个操作为以下四种操作之一:

    • + u v c:将u到v的路径上的点的权值都加上自然数c;

    • - u1 v1 u2 v2:将树中原有的边(u1,v1)删除,加入一条新边(u2,v2),保证操作完之后仍然是一棵树;

    • * u v c:将u到v的路径上的点的权值都乘上自然数c;

    • / u v:询问u到v的路径上的点的权值和,求出答案对于51061的余数。

    输入输出格式

    输入格式:

    第一行两个整数n,q

    接下来n-1行每行两个正整数u,v,描述这棵树

    接下来q行,每行描述一个操作

    输出格式:

    对于每个/对应的答案输出一行

    输入输出样例

    输入样例#1: 复制
    3 2
    1 2
    2 3
    * 1 3 4
    / 1 1
    输出样例#1: 复制
    4
    

    说明

    10%的数据保证,1<=n,q<=2000

    另外15%的数据保证,1<=n,q<=5*10^4,没有-操作,并且初始树为一条链

    另外35%的数据保证,1<=n,q<=5*10^4,没有-操作

    100%的数据保证,1<=n,q<=10^5,0<=c<=10^4

    By (伍一鸣)

    全程自己YY,调了一下午真累啊QWQ......

    LCT的板子题,要维护子树和,加法标记,乘法标记和自身的值

    放标记的时候先放乘法标记

    至于为什么http://www.cnblogs.com/zwfymqz/p/8588693.html

    // luogu-judger-enable-o2
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #define int long long 
    const int mod = 51061; 
    using namespace std;
    const int MAXN = 1e5+10;
    inline int read() {
        char c = getchar(); int x = 0, f = 1;
        while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
        while(c >= '0' && c <= '9') {x = x * 10 + c - '0'; c = getchar();}
        return x * f;
    }
    int N, Q;
    #define ls(x) T[x].ch[0]
    #define rs(x) T[x].ch[1]
    #define fa(x) T[x].f
    struct node {
        int f, ch[2], r, mul, add, siz, sum, val;
    }T[MAXN];
    bool IsRoot(int x) {
        return T[fa(x)].ch[0] != x && T[fa(x)].ch[1] != x;
    }
    int ident(int x) {
        return T[fa(x)].ch[0] == x ? 0 : 1;
    }
    void connect(int x, int fa, int how) {
        T[x].f = fa;
        T[fa].ch[how] = x;
    }
    void update(int x) {
        T[x].sum = (T[ls(x)].sum + T[rs(x)].sum + T[x].val ) % mod;
        T[x].siz = T[ls(x)].siz + T[rs(x)].siz + 1 ;
    }
    void rotate(int x) {
        int Y = fa(x), R = fa(Y), Yson = ident(x), Rson = ident(Y);
        int B = T[x].ch[Yson ^ 1];
        T[x].f = R;
        if(!IsRoot(Y)) 
            connect(x, R, Rson);
        connect(B, Y, Yson);
        connect(Y, x, Yson ^ 1);
        update(Y);update(x);
    }
    void pushr(int x) {
        if(T[x].r) {
            swap(ls(x), rs(x));
            T[ls(x)].r ^= 1;
            T[rs(x)].r ^= 1;
            T[x].r = 0;
        }
    }
    void pushmul(int x) {
        T[ls(x)].val *= T[x].mul; T[ls(x)].val %= mod;
        T[rs(x)].val *= T[x].mul; T[rs(x)].val %= mod;
        T[ls(x)].sum *= T[x].mul; T[ls(x)].sum %= mod;
        T[rs(x)].sum *= T[x].mul; T[rs(x)].sum %= mod;
        T[ls(x)].add *= T[x].mul; T[ls(x)].add %= mod;
        T[rs(x)].add *= T[x].mul; T[rs(x)].add %= mod;
        T[ls(x)].mul *= T[x].mul; T[ls(x)].mul %= mod;
        T[rs(x)].mul *= T[x].mul; T[rs(x)].mul %= mod;
        T[x].mul = 1;
    }
    void pushadd(int x) {
        T[ls(x)].val += T[x].add; T[ls(x)].val %= mod;
        T[rs(x)].val += T[x].add; T[rs(x)].val %= mod;
        T[ls(x)].sum += T[ls(x)].siz * T[x].add; T[ls(x)].sum %= mod;
        T[rs(x)].sum += T[rs(x)].siz * T[x].add; T[rs(x)].sum %= mod;
        T[ls(x)].add += T[x].add; T[ls(x)].add %= mod;
        T[rs(x)].add += T[x].add; T[rs(x)].add %= mod;
        T[x].add = 0;
    }
    void pushdown(int x) {
        pushr(x);
        pushmul(x);
        pushadd(x);
    }
    int st[MAXN];
    int fuck;
    void splay(int x) {
        int y = x, top = 0;
        st[++top] = y;
        while(!IsRoot(y)) st[++top] = (y = fa(y));
        while(top) pushdown(st[top--]);
        for(int y = fa(x); !IsRoot(x); rotate(x), y = fa(x))
            if(!IsRoot(y))
                rotate( ident(x) == ident(y) ? y : x );
    }
    void access(int x) {
        for(int y = 0; x; x = fa(y = x))//tag
            splay(x), rs(x) = y, update(x);
    }
    void makeroot(int x) {
        access(x); 
        splay(x);
        T[x].r ^= 1;
    }
    void split(int x,int y) {
        makeroot(x);
        access(y);
        splay(y);
    }
    void link(int x, int y) {
        makeroot(x);
        T[x].f = y;
    }
    int findroot(int x) {
        access(x); 
        splay(x);
        while(ls(x)) x = ls(x);
        return x;
    }
    void cut(int x, int y) {
        split(x, y);
        //makeroot(x);
        if(findroot(y) == x && fa(x) == y && !rs(x))
            T[x].f = T[y].ch[0] = 0, update(y);
    }
    main() {
        freopen("a.in","r",stdin);
        scanf("%lld%lld", &N, &Q);
        for(int i = 1; i <= N; i++) 
            T[i].val = 1, T[i].sum = 1, T[i].mul = 1, T[i].siz = 1, T[i].add = 0;
        for(int i = 1; i <= N - 1; i++) {
            int x = read(), y = read();
            link(x, y);
        }
        int fuck = 0;
        while(Q--) {
            char c = 0; int u1, v1, u2, v2, val;
            while(c < '*') c = getchar(); 
            scanf("%lld%lld", &u1, &v1);
            if(c == '+') {
                scanf("%lld", &val);
                split(u1, v1);
                T[v1].val += val; T[v1].val %= mod;
                T[v1].add += val; T[v1].add %= mod;
                T[v1].sum += T[v1].siz * val; T[v1].sum %= mod;
            }    
            else if(c == '-') {
                scanf("%lld%lld", &u2, &v2);
                cut(u1, v1);
                link(u2, v2);
            }
            else if(c == '/') {
                split(u1, v1);
                printf("%d
    ",T[v1].sum%mod);            
            }
            else {
                scanf("%lld",&val);
                split(u1, v1);
                T[v1].val *= val; T[v1].val %= mod;
                T[v1].sum *= val; T[v1].siz %= mod;
                T[v1].add *= val; T[v1].add %= mod;
                T[v1].mul *= val; T[v1].mul %= mod;
            }    
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/zwfymqz/p/8971356.html
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