这个题和POJ 1741类似。。。。只不过求的是在树上是否存在路径长度等于K的路径。。。解法也是类似的。。。就在分治中求出路径长度等于k的路径个数,如果是根节点就加上,儿子节点就减去。总复杂度m*n*log(n)*log(n)
#include <iostream>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 10005
#define maxm 20005
#define eps 1e-7
#define mod 1000000007
#define INF 0x3f3f3f3f
#define PI (acos(-1.0))
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid
#define rson o<<1 | 1, mid+1, R
#define pii pair<int, int>
#pragma comment(linker, "/STACK:16777216")
typedef long long LL;
typedef unsigned long long ULL;
//typedef int LL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
//head
struct Edge
{
int v, w;
Edge *next;
}*H[maxn], E[maxm], *edges;
vector<int> dis;
bool done[maxn];
int size[maxn];
int mx[maxn];
int a[105];
int b[105];
int root, n, m, nsize;
void addedges(int u, int v, int w)
{
edges->v = v;
edges->w = w;
edges->next = H[u];
H[u] = edges++;
}
void init()
{
edges = E;
memset(b, 0, sizeof b);
memset(H, 0, sizeof H);
memset(done, 0, sizeof done);
}
void read()
{
int v, w;
for(int i = 1; i <= n; i++) {
scanf("%d", &v);
if(v == 0) continue;
scanf("%d", &w);
addedges(i, v, w);
addedges(v, i, w);
i--;
}
m = 0;
while(scanf("%d", &w), w != 0) a[++m] = w;
}
void getroot(int u, int fa)
{
size[u] = 1, mx[u] = 0;
for(Edge *e = H[u]; e; e = e->next) if(!done[e->v] && e->v != fa) {
int v = e->v;
getroot(v, u);
size[u] += size[v];
mx[u] = max(mx[u], size[v]);
}
mx[u] = max(mx[u], nsize - size[u]);
if(mx[u] < mx[root]) root = u;
}
void getdis(int u, int fa, int dist)
{
dis.push_back(dist);
for(Edge *e = H[u]; e; e = e->next) if(!done[e->v] && fa != e->v) {
int v = e->v, w = e->w;
getdis(v, u, dist + w);
}
}
void calc(int u, int fa, int dist, int flag)
{
dis.clear();
getdis(u, fa, dist);
sort(dis.begin(), dis.end());
for(int i = 1; i <= m; i++) {
for(int l = 0, r = dis.size() - 1; l < r; ) {
if(dis[l] + dis[r] == a[i]) {
b[i] += flag;
if(r && dis[r-1] == dis[r]) r--;
else l++;
}
else if(dis[l] + dis[r] > a[i]) r--;
else l++;
}
}
}
void solve(int u)
{
calc(u, 0, 0, 1);
done[u] = true;
for(Edge *e = H[u]; e; e = e->next) if(!done[e->v]) {
int v = e->v, w = e->w;
calc(v, u, w, -1);
mx[0] = nsize = size[v];
getroot(v, root = 0);
solve(root);
}
}
void work()
{
mx[0] = nsize = n;
getroot(1, root = 0);
solve(root);
for(int i = 1; i <= m; i++)
if(b[i]) printf("AYE\n");
else printf("NAY\n");
printf(".\n");
}
int main()
{
while(scanf("%d", &n), n != 0) {
init();
read();
work();
}
return 0;
}