BZOJ 1316: 树上的询问( 点分治 + 平衡树 )

直接点分治, 用平衡树(set就行了...)维护. 

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#include<cstdio>

#include<cstring>

#include<algorithm>

#include<cctype>

#include<set>

 

using namespace std;

 

typedef long long ll;

 

const int maxn = 10009;

const int maxm = 109;

 

int read() {

char c = getchar();

int ret = 0;

for(; !isdigit(c); c = getchar());

for(; isdigit(c); c = getchar()) ret = ret * 10 + c - '0';

return ret;

}

 

struct edge {

int to, w;

edge* next;

} E[maxn << 1], *pt = E, *head[maxn];

 

void AddEdge(int u, int v, int w) {

pt->to = v; pt->w = w; pt->next = head[u]; head[u] = pt++;

}

 

int n, Rt, Min, size[maxn];

int N, qn;

ll q[maxm];

bool ans[maxm], vis[maxn];

set<ll> Bst;

 

void init() {

N = read(); qn = read();

for(int i = 1; i < N; i++) {

int u = read() - 1, v = read() - 1, w = read();

AddEdge(u, v, w);

AddEdge(v, u, w);

}

for(int i = 0; i < qn; i++)

scanf("%lld", q + i);

n = N;

}

 

void dfs(int x, int fa) {

size[x] = 1;

int Max = 0;

for(edge* e = head[x]; e; e = e->next) if(e->to != fa && !vis[e->to]) {

dfs(e->to, x);

size[x] += size[e->to];

Max = max(size[e->to], Max);

}

Max = max(Max, n - size[x]);

if(Max < Min)

Min = Max, Rt = x;

}

 

void Insert(int x, int fa, ll w) {

Bst.insert(w);

for(edge* e = head[x]; e; e = e->next) if(!vis[e->to] && fa != e->to)

Insert(e->to, x, w + e->w);

}

 

void update(int x, int fa, ll w) {

for(int i = 0; i < qn; i++) if(!ans[i])

ans[i] |= (Bst.find(q[i] - w) != Bst.end());

for(edge* e = head[x]; e; e = e->next) if(!vis[e->to] && fa != e->to)

update(e->to, x, w + e->w);

}

 

void solve(int x) {

Min = maxn;

dfs(x, -1);

x = Rt;

Bst.clear();

Bst.insert(0);

for(edge* e = head[x]; e; e = e->next) if(!vis[e->to]) {

update(e->to, x, e->w);

Insert(e->to, x, e->w);

}

vis[x] = true;

for(edge* e = head[x]; e; e = e->next) if(!vis[e->to]) {

n = size[e->to];

solve(e->to);

}

}

 

int main() {

init();

memset(vis, 0, sizeof vis);

memset(ans, 0, sizeof ans);

solve(0);

for(int i = 0; i < qn; i++)

if(!q[i]) ans[i] = true;

for(int i = 0; i < qn; i++)

puts(ans[i] ? "Yes" : "No");

return 0;

}

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1316: 树上的询问

Time Limit: 10 Sec  Memory Limit: 162 MB
Submit: 417  Solved: 103
[Submit][Status][Discuss]

Description

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.

Input

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．

Output

输出有p行，Yes或No.

Sample Input

6 4
1 2 5
1 3 7
1 4 1
3 5 2
3 6 3
1
8
13
14

Sample Output

Yes
Yes
No
Yes

HINT

30%的数据，n≤100． 
100%的数据，n≤10000，p≤100，长度≤1000000． 

做完此题可看下POJ 3237 Tree

Source