• Tree 点分治


    题目描述

    给你一棵TREE,以及这棵树上边的距离.问有多少对点它们两者间的距离小于等于K

    输入输出格式

    输入格式:

    N(n<=40000) 接下来n-1行边描述管道,按照题目中写的输入 接下来是k

    输出格式:

    一行,有多少对点之间的距离小于等于k

    输入输出样例

    输入样例#1: 
    7
    1 6 13 
    6 3 9 
    3 5 7 
    4 1 3 
    2 4 20 
    4 7 2 
    10
    输出样例#1: 
    5





    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    using namespace std;
    inline int read() {
        int res=0;char ch=getchar();
        while(!isdigit(ch)) ch=getchar();
        while(isdigit(ch)) res=(res<<3)+(res<<1)+(ch^48),ch=getchar();
        return res;
    }
    #define reg register
    #define N 200005
    int n, tot, k;
    long long ans;
    
    int head[N], cnt = 1;
    struct edge {
        int nxt, to, val;
    }ed[N*2];
    inline void add(int x, int y, int z)
    {
        ed[++cnt] = (edge){head[x], y, z};
        head[x] = cnt;
    }
    
    bool cut[N*2];
    
    
    int siz[N], root, mrt = 1e9;
    void dfs(int x, int fa)
    {
        siz[x] = 1;
        for (reg int i = head[x] ; i ; i = ed[i].nxt)
        {
            int to = ed[i].to;
            if (to == fa or cut[i]) continue;
            dfs(to, x);
            siz[x] += siz[to];
        }    
    }    
    void efs(int x, int fa)
    {
        int tmp = tot - siz[x];
        for (reg int i = head[x] ; i ; i = ed[i].nxt)
        {
            int to = ed[i].to;
            if (to == fa or cut[i]) continue;
            efs(to, x);
            tmp = max(tmp, siz[to]);
        }
        if (tmp < mrt) mrt = tmp, root = x;
    }    
    inline int FindRoot(int x)
    {
        return x;
        dfs(x, 0);
        mrt = 1e9;
        tot = siz[x];
        root = n;
        efs(x, 0);
        return root;
    }
    
    
    int a[N];
    int top;
    void Work(int x, int fa, int d)
    {
        a[++top] = d;
        for (reg int i = head[x] ; i ; i = ed[i].nxt)
        {
            int to = ed[i].to;
            if (to == fa or cut[i]) continue;
            Work(to, x, d + ed[i].val);
        }
    }
    inline int Calc(int x, int d)
    {
        int res = 0;
        top = 0;
        Work(x, 0, d);
        sort (a + 1, a + 1 + top);
        int l = 1, r = top;
        while (l < r)
        {
            while(a[l] + a[r] > k and l < r) r--;
            res += r - l, l ++;
        }        
        return res;
    }
    void solve(int rt)
    {
        root = FindRoot(rt);
        ans += Calc(root, 0);
        for (reg int i = head[root] ; i ; i = ed[i].nxt)
        {
            int to = ed[i].to;
            if (cut[i]) continue;
            cut[i] = cut[i ^ 1] = 1;
            ans -= Calc(to, ed[i].val);
            solve(to);
        }
    }
        
    
    int main()
    {
        n = read();
        for (reg int i = 1 ; i < n ; i ++)
        {
            int x = read(), y = read(), z = read();
            add(x, y, z), add(y, x, z);
        } k = read();
        solve(1);
        printf("%d
    ", ans);
        return 0;
    }


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  • 原文地址:https://www.cnblogs.com/BriMon/p/9479963.html
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