BZOJ2599 [IOI2011]Race(点分治)

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题目描述:

给一棵树,每条边有权.求一条路径,权值和等于K,且边的数量最小.

输入描述:

第一行 两个整数 n, k
第二…n行 每行三个整数 表示一条无向边的两端和权值 (注意点的编号从0开始)
(n≤200000,K≤1000000)

输出描述:

一个整数 表示最小边数量 如果不存在这样的路径 输出-1

样例输入:

4 3
0 1 1
1 2 2
1 3 4

样例输出:

2

思路

利用点分治

开一个100W的数组t,t[i]表示权值为i的路径最少边数
找到重心分成若干子树后, 得出一棵子树的所有点到根的权值和x,到根a条边,用t[k-x]+a更新答案,全部查询完后,再用所有a更新t[x],这样可以保证不出现点分治中的不合法情况

d[i]当前树走到i点走过的边数,deep[i]表示走过的路径

代码

#include <bits/stdc++.h>
using namespace std;
#define mem(a, b) memset(a, b, sizeof(a))
const int inf = 0x3f3f3f3f;
const int N = 2e5 + 10;
const int M = 1e6 + 10;
int n, k;
int first[N], tot;
int f[N], siz[N], vis[N], ans, sum;
int d[N], deep[N]; //d[i]当前树走到i点走过的边数,deep[i]表示走过的路径
int tmp[M], root;
struct edge
{
    int v, w, next;
} e[N * 2];
void add_edge(int u, int v, int w)
{
    e[tot].v = v, e[tot].w = w;
    e[tot].next = first[u];
    first[u] = tot++;
}
void init()
{
    mem(first, -1);
    tot = 0;
    root = 0;
    sum = f[0] = n;
    ans = inf;
}
void getroot(int u, int fa)
{
    siz[u] = 1, f[u] = 0;
    for (int i = first[u]; ~i; i = e[i].next)
    {
        int v = e[i].v;
        if (v == fa || vis[v])
            continue;
        getroot(v, u);
        siz[u] += siz[v];
        f[u] = max(f[u], siz[v]);
    }
    f[u] = max(f[u], sum - siz[u]);
    if (f[u] < f[root])
        root = u;
}
void getdeep(int u, int fa)
{
    if (deep[u] <= k)
        ans = min(ans, tmp[k - deep[u]] + d[u]);
    for (int i = first[u]; ~i; i = e[i].next)
    {
        int v = e[i].v, w = e[i].w;
        if (v == fa || vis[v])
            continue;
        deep[v] = deep[u] + w;
        d[v] = d[u] + 1;
        getdeep(v, u);
    }
}
void update(int u, int fa, int flag)
{
    if (deep[u] <= k)
    {
        if (flag)
            tmp[deep[u]] = min(tmp[deep[u]], d[u]);
        else
            tmp[deep[u]] = inf;
    }
    for (int i = first[u]; ~i; i = e[i].next)
    {
        int v = e[i].v;
        if (v == fa || vis[v])
            continue;
        update(v, u, flag);
    }
}
void solve(int u)
{
    vis[u] = 1;
    tmp[0] = 0;
    for (int i = first[u]; ~i; i = e[i].next)
    {
        int v = e[i].v, w = e[i].w;
        if (vis[v])
            continue;
        deep[v] = w;
        d[v] = 1;
        getdeep(v, 0);
        update(v, 0, 1);
    }
    for (int i = first[u]; ~i; i = e[i].next)
    {
        int v = e[i].v;
        if (!vis[v])
            update(v, 0, 0);
    }
    for (int i = first[u]; ~i; i = e[i].next)
    {
        int v = e[i].v;
        if (vis[v])
            continue;
        sum = siz[v];
        root = 0;
        getroot(v, 0);
        solve(root);
    }
}
int main()
{
    int u, v, w;
    scanf("%d%d", &n, &k);
    init();
    for (int i = 1; i <= k; i++)
        tmp[i] = inf;
    for (int i = 1; i <= n - 1; i++)
    {
        scanf("%d%d%d", &u, &v, &w);
        u++, v++;
        add_edge(u, v, w);
        add_edge(v, u, w);
    }
    getroot(1, 0);
    solve(root);
    if (ans != inf)
        printf("%d\n", ans);
    else
        printf("-1");

    return 0;
}

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