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题目描述:
一棵n个点的带权有根树，有p个询问，每次询问树中是否存在一条长度为Len的路径，如果是，输出Yes否输出No.
输入描述:
第一行两个整数n, p分别表示点的个数和询问的个数． 接下来n-1行每行三个数x, y, c，表示有一条树边x→y，长度为c． 接下来p行每行一个数Len，表示询问树中是否存在一条长度为Len的路径．
输出描述:
输出有p行，Yes或No.
样例输入:
6 4
1 2 5
1 3 7
1 4 1
3 5 2
3 6 3
1
8
13
14
样例输出:
Yes
Yes
No
Yes
提示:
30%的数据，n≤100．
100%的数据，n≤10000，p≤100，长度≤1000000．
思路
利用点分治，首先把树的重心找出来之后，然后以这一点为根，递归的解决,点分治处理的是经过根节点的路径，所以在计算完当前节点之后要去重(儿子)，每次把处理的所有深度排序，然后遍历每一个深度值，看一下存在k-x的有多少个累加这个值就是答案，利用二分实现。先把所有的查询存下来，然后离线处理。
代码
#include <bits/stdc++.h>
using namespace std;
#define mem(a, b) memset(a, b, sizeof(a))
const int N = 1e4 + 10;
int first[N], tot, n, p;
int q[N], siz[N], f[N], d[N], deep[N], vis[N];
int sum, 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;
sum = f[0] = n;
root = 0;
}
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 (!vis[v] && v != fa)
{
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)
{
deep[++deep[0]] = d[u];
for (int i = first[u]; ~i; i = e[i].next)
{
int v = e[i].v, w = e[i].w;
if (!vis[v] && v != fa)
{
d[v] = d[u] + w;
getdeep(v, u);
}
}
}
int findl(int l, int r, int k)
{
int ans = 0;
while (l <= r)
{
int mid = (l + r) >> 1;
if (deep[mid] == k)
{
ans = mid;
r = mid - 1;
}
else if (deep[mid] < k)
l = mid + 1;
else
r = mid - 1;
}
return ans;
}
int findr(int l, int r, int k)
{
int ans = -1;
while (l <= r)
{
int mid = (l + r) >> 1;
if (deep[mid] == k)
{
ans = mid;
l = mid + 1;
}
else if (deep[mid] < k)
l = mid + 1;
else
r = mid - 1;
}
return ans;
}
int cal(int u, int cost, int k)
{
d[u] = cost;
deep[0] = 0;
getdeep(u, 0);
sort(deep + 1, deep + deep[0] + 1);
int t = 0;
for (int i = 1; i <= deep[0]; i++)
{
if (deep[i] + deep[i] > k)
break;
int l = findl(i, deep[0], k - deep[i]);
int r = findr(i, deep[0], k - deep[i]);
/*二分的部分也可以用lower_bound和upper_bound实现
int l = lower_bound(deep + 1, deep + deep[0] + 1, k - deep[i]) - deep;
int r = upper_bound(deep + 1, deep + deep[0] + 1, k - deep[i]) - deep - 1;
*/
t += r - l + 1;
}
return t;
}
int ans[110];
void solve(int u)
{
for (int i = 1; i <= p; i++)
ans[i] += cal(u, 0, q[i]);
vis[u] = 1;
for (int i = first[u]; ~i; i = e[i].next)
{
int v = e[i].v, w = e[i].w;
if (!vis[v])
{
for (int j = 1; j <= p; j++)
ans[j] -= cal(v, w, q[j]);
sum = siz[v];
root = 0;
getroot(v, 0);
solve(root);
}
}
}
int main()
{
// freopen("in.txt", "r", stdin);
int u, v, w;
scanf("%d%d", &n, &p);
init();
for (int i = 1; i <= n - 1; i++)
{
scanf("%d%d%d", &u, &v, &w);
add_edge(u, v, w);
add_edge(v, u, w);
}
for (int i = 1; i <= p; i++)
scanf("%d", &q[i]);
getroot(1, 0);
solve(root);
for (int i = 1; i <= p; i++)
puts(ans[i] ? "Yes" : "No");
return 0;
}