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  • [SHOI 2012] 魔法树

    [题目链接]

            https://www.lydsy.com/JudgeOnline/problem.php?id=2836

    [算法]

           树链剖分

           时间复杂度 : O(NlogN ^ 2)

    [代码]

           

    #include<bits/stdc++.h>
    using namespace std;
    #define MAXN 100010
    typedef long long LL;
    
    struct edge
    {
        int to , nxt;
    } e[MAXN << 1];
    
    int n , tot , timer;
    int head[MAXN] , size[MAXN] , son[MAXN]  , fa[MAXN] , dfn[MAXN] , depth[MAXN] , top[MAXN];
    
    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;
    }
    struct Segment_Tree
    {
        struct Node
        {
            int l , r;
            LL sum , tag;
        } Tree[MAXN << 2];
        inline void build(int index , int l , int r)
        {
            Tree[index].l = l;
            Tree[index].r = r;
            Tree[index].tag = Tree[index].sum = 0;
            if (l == r) return;
            int mid = (l + r) >> 1;
            build(index << 1 , l , mid);
            build(index << 1 | 1 , mid + 1 , r);
        }
        inline void pushdown(int index)
        {
            int l = Tree[index].l , r = Tree[index].r;
            int mid = (l + r) >> 1;
            Tree[index << 1].sum += (mid - l + 1) * Tree[index].tag;
            Tree[index << 1 | 1].sum += (r - mid) * Tree[index].tag;
            Tree[index << 1].tag += Tree[index].tag;
            Tree[index << 1 | 1].tag += Tree[index].tag;
            Tree[index].tag = 0;
        }
        inline void update(int index)
        {
            Tree[index].sum = Tree[index << 1].sum + Tree[index << 1 | 1].sum;
        }
        inline void modify(int index , int l , int r , LL value)
        {
            if (Tree[index].l == l && Tree[index].r == r)
            {
                Tree[index].sum += value * (r - l + 1);
                Tree[index].tag += value;
                return;
            }
            pushdown(index);
            int mid = (Tree[index].l + Tree[index].r) >> 1;
            if (mid >= r) modify(index << 1 , l , r , value);
            else if (mid + 1 <= l) modify(index << 1 | 1 , l , r , value);
            else
            {
                modify(index << 1 , l , mid , value);
                modify(index << 1 | 1 , mid + 1 , r , value);
            } 
            update(index); 
        }
        inline LL query(int index , int l , int r)
        {
            if (Tree[index].l == l && Tree[index].r == r) return Tree[index].sum;
            pushdown(index);
            int mid = (Tree[index].l + Tree[index].r) >> 1;
            if (mid >= r) return query(index << 1 , l , r);
            else if (mid + 1 <= l) return query(index << 1 | 1 , l , r);
            else return query(index << 1 , l , mid) + query(index << 1 | 1 , mid + 1 , r);
        }
    } SGT;
    inline void addedge(int u , int v)
    {
        ++tot;
        e[tot] = (edge){v , head[u]};
        head[u] = tot;
    } 
    inline void dfs1(int u)
    {
        son[u] = -1;
        size[u] = 1;
        for (int i = head[u]; i; i = e[i].nxt)
        {
            int v = e[i].to;
            if (v == fa[u]) continue;
            fa[v] = u;
            depth[v] = depth[u] + 1;
            dfs1(v);
            size[u] += size[v];
            if (son[u] == -1 || size[v] > size[son[u]]) son[u] = v;
        }
    }
    inline void dfs2(int u , int tp)
    {
        dfn[u] = ++timer;
        top[u] = tp;
        if (son[u] != -1) dfs2(son[u] , tp);
        for (int i = head[u]; i; i = e[i].nxt)
        {
            int v = e[i].to;
            if (v == fa[u] || v == son[u]) continue;
            dfs2(v , v);
        }
    }
    inline void modify(int u , int v , LL d)
    {
        int tu = top[u] , tv = top[v];
        while (tu != tv)
        {
            if (depth[tu] > depth[tv])
            {
                swap(u , v);
                swap(tu , tv);
            }    
            SGT.modify(1 , dfn[tv] , dfn[v] , d);
            v = fa[tv]; tv = top[v];
        }    
        if (depth[u] > depth[v]) swap(u , v);
        SGT.modify(1, dfn[u] , dfn[v] , d);
    }
    
    int main()
    {
        
        scanf("%d" , &n);
        for (int i = 1; i < n; i++)
        {
            int u , v;
            scanf("%d%d" , &u , &v);
            addedge(u , v);    
            fa[v] = u;
        }
        dfs1(0);
        dfs2(0 , 0);
        SGT.build(1 , 1 , n);
        int q;
        scanf("%d" , &q);
        while (q--)
        {
            char op[5];
            scanf("%s" , &op);
            if (op[0] == 'A')
            {
                int u , v;
                LL d;
                scanf("%d%d%lld" , &u , &v , &d);
                modify(u , v , d);
            } else
            {
                int u;
                scanf("%d" , &u);
                printf("%lld
    " , SGT.query(1 , dfn[u] , dfn[u] + size[u] - 1));
            }
        }
        
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/evenbao/p/9908582.html
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