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[POI2007] 大都市

时间:2018-11-05 14:02:11      阅读:247      评论:0      收藏:0      [点我收藏+]

标签:color   树链剖分   max   def   while   online   for   复杂   oid   

[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\n" , query(u));
        }
    }
    
    return 0;
}

 

[POI2007] 大都市

标签:color   树链剖分   max   def   while   online   for   复杂   oid   

原文地址:https://www.cnblogs.com/evenbao/p/9908708.html

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