• [POI2007] 大都市


    [题目链接]

               https://www.lydsy.com/JudgeOnline/problem.php?id=1103

    [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 250010
    typedef long long LL;
    
    int n , tot , timer;
    int head[MAXN] , dfn[MAXN] , top[MAXN] , fa[MAXN] , size[MAXN] , son[MAXN] , depth[MAXN];
    
    struct edge
    {
        int to , nxt;
    } e[MAXN << 1];
    
    struct Segment_Tree
    {
        struct Node
        {
            int l , r;
            int sum , tag;
        } Tree[MAXN << 2];
        inline void build(int index , int l , int r)
        {
            Tree[index] = (Node){l , r , r - l + 1 , 0};
            if (l == 1) --Tree[index].sum;
            if (l == r) return;
            int mid = (Tree[index].l + Tree[index].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;
            if (Tree[index].tag)
            {
                Tree[index << 1].sum = Tree[index << 1 | 1].sum = 0;
                Tree[index << 1].tag = Tree[index << 1 | 1].tag = 1;
            }
        }
        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)
        {
            if (l > r) return;
            if (Tree[index].l == l && Tree[index].r == r) 
            {
                Tree[index].sum = 0;
                Tree[index].tag = 1;
                return;
            }
            pushdown(index);
            int mid = (Tree[index].l + Tree[index].r) >> 1;
            if (mid >= r) modify(index << 1 , l , r);
            else if (mid + 1 <= l) modify(index << 1 | 1 , l , r);
            else
            {
                modify(index << 1 , l , mid);
                modify(index << 1 | 1 , mid + 1 , r);
            }
            update(index);
        }
        inline int query(int index , int l , int r)
        {
            if (l > r) return 0;
            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;
    
    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 void addedge(int u , int v)
    {
        ++tot;
        e[tot] = (edge){v , head[u]};
        head[u] = tot; 
    }
    inline void dfs1(int u)
    {
        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 (size[v] > size[son[u]]) son[u] = v;    
        }    
    }
    inline void dfs2(int u , int tp)
    {
        dfn[u] = ++timer;
        top[u] = tp;
        if (son[u]) 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)
    {
        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]);
            v = fa[tv]; tv = top[v];
        }    
        if (depth[u] > depth[v]) swap(u , v);
        SGT.modify(1 , dfn[u] + 1 , dfn[v]);
    }
    inline int query(int u)
    {
        int tu = top[u] , ret = 0;
        while (tu != 1)
        {
            ret    += SGT.query(1 , dfn[tu] , dfn[u]);
            u = fa[tu]; tu = top[u];
        }    
        ret += SGT.query(1 , 1 , dfn[u]);
        return ret;
    }
    
    int main()
    {
        
        scanf("%d" , &n);
        for (int i = 1; i < n; i++)
        {
            int u , v;
            scanf("%d%d" , &u , &v);
            addedge(u , v);    
            addedge(v , u);
        }
        dfs1(1);
        dfs2(1 , 1);
        SGT.build(1 , 1 , n);
        int m;
        scanf("%d" , &m);
        for (int i = 1; i <= n + m - 1; i++)
        {
            char op[5];
            scanf("%s" , &op);
            if (op[0] == 'A')
            {
                int u , v;
                scanf("%d%d" , &u , &v);
                modify(u , v);    
            } else 
            {
                int u;
                scanf("%d" , &u);
                printf("%d
    " , query(u));
            }
        }
        
        return 0;
    }
  • 相关阅读:
    地铁项目结对编程
    地铁项目初步计划及简单设计
    构建之法浅读感想
    集美大学1511,1512软件工程课程作业总结
    第二次作业小结
    第二次作业评分可能要 delay 一些
    第一次作业小结
    关于我
    面向对象设计与构造第四单元总结及期终总结
    面向对象设计与构造第三单元作业总结
  • 原文地址:https://www.cnblogs.com/evenbao/p/9908708.html
Copyright © 2020-2023  润新知