bzoj2599 [IOI2011]Race

2599: [IOI2011]Race

Time Limit: 70 Sec  Memory Limit: 128 MB
Submit: 4559  Solved: 1343
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Description

给一棵树,每条边有权.求一条简单路径,权值和等于K,且边的数量最小.N <= 200000, K <= 1000000

Input

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)

Output

一个整数 表示最小边数量 如果不存在这样的路径 输出-1

Sample Input

4 3
0 1 1
1 2 2
1 3 4

Sample Output

2

HINT

 2018.1.3新加数据一组，未重测

分析：复习了一下点分治. 这道题在点分治模板的基础上改改就好了. dfs到一个点添加重心到它的距离和经过的边数，最后两个指针扫一下就好了.

#include <cstdio>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int maxn = 300010;
int n,k,head[maxn],to[maxn * 2],nextt[maxn * 2],w[maxn * 2],tot = 1,d[maxn],num[maxn];
int sizee[maxn],f[maxn],sum,root,vis[maxn],ans[1000010],anss = -1,top;

struct node
{
    int p1,p2;
}e[maxn];

void add(int x,int y,int z)
{
    w[tot] = z;
    to[tot] = y;
    nextt[tot] = head[x];
    head[x] = tot++;
}

void findroot(int u,int fa)
{
    sizee[u] = 1;
    f[u] = 0;
    for (int i = head[u]; i; i = nextt[i])
    {
        int v = to[i];
        if (vis[v] || v == fa)
            continue;
        findroot(v,u);
        sizee[u] += sizee[v];
        f[u] = max(f[u],sizee[v]);
    }
    f[u] = max(f[u],sum - sizee[u]);
    if (f[u] < f[root])
        root = u;
}

void dfs1(int u,int d1,int d2,int fa)
{
    d[u] = d1;
    num[u] = d2;
    for (int i = head[u];i;i = nextt[i])
    {
        int v = to[i];
        if (vis[v] || v == fa)
            continue;
        dfs1(v,d1 + w[i],d2 + 1,u);
    }
}

void dfs2(int u,int fa)
{
    e[++top].p1 = d[u];
    e[top].p2 = num[u];
    for (int i = head[u];i;i = nextt[i])
    {
        int v = to[i];
        if (vis[v] || v == fa)
            continue;
        dfs2(v,u);
    }
}

bool cmp(node a,node b)
{
    if (a.p1 == b.p1)
        return a.p2 < b.p2;
    return a.p1 < b.p1;
}

void calc1(int u,int W)
{
    top = 0;
    dfs2(u,0);
    sort(e + 1,e + 1 + top,cmp);
    int l = 1,r = top;
    while (l <= r)
    {
        while (l < r && e[l].p1 + e[r].p1 > k)
            r--;
        for (int kk = r; e[l].p1 + e[kk].p1 == k; kk--)  //这里不是直接移动r.
            ans[e[kk].p2 + e[l].p2]++;
        l++;
    }
}

void calc2(int u,int W)
{
    top = 0;
    dfs2(u,0);
    sort(e + 1,e + 1 + top,cmp);
    int l = 1,r = top;
    while (l <= r)
    {
        while (l < r && e[l].p1 + e[r].p1 > k)
            r--;
        for (int kk = r; e[l].p1 + e[kk].p1 == k; kk--)
            ans[e[kk].p2 + e[l].p2]--;
        l++;
    }
}

void solve(int u)
{
    vis[u] = 1;
    dfs1(u,0,0,0);
    calc1(u,0);
    for (int i = head[u]; i; i = nextt[i])
    {
        int v = to[i];
        if (vis[v])
            continue;
        calc2(v,w[i]);
        f[0] = sum = sizee[v];
        findroot(v,root = 0);
        solve(root);
    }
}

int main()
{
    scanf("%d%d",&n,&k);
    for (int i = 1; i < n; i++)
    {
        int x,y,z;
        scanf("%d%d%d",&x,&y,&z);
        x++;
        y++;
        add(x,y,z);
        add(y,x,z);
    }
    f[0] = sum = n;
    findroot(1,0);
    solve(root);
    for (int i = 1; i <= n; i++)
        if (ans[i])
        {
            anss = i;
            break;
        }
    printf("%d
",anss);

    return 0;
}