[BZOJ1468]Tree 题解

[BZOJ1468]Tree 题解

题目地址:洛谷:【P4178】Tree – 洛谷、BZOJ:Problem 1468. — Tree

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

题解

点分治的板子。下面只说统计答案的思路。
我们考虑每一对儿子,计算其经过根的路径。如果按照【P3806】【模板】点分治1 – 洛谷这道题O(n^2)地统计答案可能有些吃力,我们要把统计答案改为O(n)算法,即维护两个指针l和r来拼出答案。但是这样计算会有不合法的情况,即两个儿子在同一子树内。那么我们考虑递归到子树里面去把这些不合法的答案减掉。递归下去以后还要把所有儿子的距离加上子树-根这条边的边权的2倍。

代码

// Code by KSkun, 2018/3
#include <cstdio>

#include <vector>
#include <algorithm>

typedef long long LL;

inline char fgc() {
    static char buf[100000], *p1 = buf, *p2 = buf;
    return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? EOF : *p1++;
}

inline LL readint() {
    register LL res = 0, neg = 1;
    char c = fgc();
    while(c < '0' || c > '9') {
        if(c == '-') neg = -1;
        c = fgc();
    }
    while(c >= '0' && c <= '9') {
        res = res * 10 + c - '0';
        c = fgc();
    }
    return res * neg;
}

const int MAXN = 40005, INF = 1e9;

int n, m, ut, vt, wt, k, rt, res = 0;
int siz[MAXN], dep[MAXN], msz[MAXN], sum;
bool vis[MAXN];

struct Edge {
    int to, w;
};

std::vector<Edge> gra[MAXN];

int ans[MAXN], atot = 0;

inline void getroot(int u, int fa) {
    siz[u] = 1; msz[u] = 0;
    for(int i = 0; i < gra[u].size(); i++) {
        int v = gra[u][i].to;
        if(vis[v] || v == fa) continue;
        getroot(v, u);
        siz[u] += siz[v];
        msz[u] = std::max(msz[u], siz[v]);
    }
    msz[u] = std::max(msz[u], sum - siz[u]);
    if(msz[u] < msz[rt]) rt = u;
}

inline void caldep(int u, int fa) {
    ans[atot++] = dep[u];
    for(int i = 0; i < gra[u].size(); i++) {
        int v = gra[u][i].to, w = gra[u][i].w;
        if(vis[v] || v == fa) continue;
        dep[v] = dep[u] + w;
        caldep(v, u); 
    }
}

inline int work(int u, int w) {
    atot = 0;
    dep[u] = w;
    caldep(u, 0);
    std::sort(ans, ans + atot);
    int l = 0, r = atot - 1;
    int resw = 0;
    while(l < r) {
        if(ans[l] + ans[r] <= k) {
            resw += r - l;
            l++;
        } else {
            r--;
        }
    }
    return resw;
}

inline void dfs(int u) {
    res += work(u, 0);
    vis[u] = true;
    for(int i = 0; i < gra[u].size(); i++) {
        int v = gra[u][i].to, w = gra[u][i].w;
        if(vis[v]) continue;
        res -= work(v, w);
        rt = 0;
        sum = siz[v];
        getroot(v, 0);
        dfs(rt);
    }
}

int main() {
    n = readint();
    for(int i = 1; i < n; i++) {
        ut = readint(); vt = readint(); wt = readint();
        gra[ut].push_back(Edge {vt, wt});
        gra[vt].push_back(Edge {ut, wt});
    }
    k = readint();
    rt = 0;
    msz[0] = INF;
    sum = n;
    getroot(1, 0);
    dfs(rt);
    printf("%d", res);
    return 0;
}


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