Tree of Three


Time Limit: 2 Seconds      Memory Limit: 65536 KB

Now we have a tree and some queries to deal with. Every node in the tree has a value on it. For one node A, we want to know the largest three values in all the nodes of the subtree whose root is node A. Node 0 is root of the tree, except it, all other nodes have a parent node.

Input

There are several test cases. Each test case begins with a line contains one integer n(1 ≤ n ≤ 10000), which indicates the number of the node in the tree. The second line contains one integer v[0], the value on the root. Then for the following n - 1 lines(from the 3rd line to the (n + 1)th line), let i + 2 be the line number, then line i + 2contains two integers parent and v[i], here parent is node i's parent node, v[i] is the value on node i. Here 0 ≤ v[i] ≤ 1000000. Then the next line contains an integer m(1 ≤m ≤ 10000), which indicates the number of queries. Following m lines, each line contains one integer q, 0 ≤ q < n, it meas a query on node q.

Output

For each test case, output m lines, each line contains one or three integers. If the query asked for a node that has less than three nodes in the subtree, output a "-1"; otherwise, output the largest three values in the subtree, from larger to smaller.

Sample Input

5
1
0 10
0 5
2 7
2 8
5
0
1
2
3
4

Sample Output

10 8 7
-1
8 7 5
-1
-1

思路:深度优先搜索,一层一层由子节点向跟节点回溯。

1.

 #include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#define MAX 11111
using namespace std;
int Three_Max[MAX][], val[MAX], cnt[MAX];
typedef struct{
int to, next;
}Node;
Node edge[MAX];
int head[MAX];
void AddEdge(int u, int v, int i){
edge[i].to = v;
edge[i].next = head[u];
head[u] = i;
}
bool cmp(int a, int b){
return a > b;
}
void dfs(int id){
cnt[id] = ;
Three_Max[id][] = val[id];
for(int i = head[id];i != -;i = edge[i].next){
int u = edge[i].to;
dfs(u);
for(int j = ;j <= ;j ++) Three_Max[id][j] = Three_Max[u][j-];
sort(Three_Max[id] + , Three_Max[id]+, cmp);
cnt[id] += cnt[u];
}
}
int main(){
int n, m, cc, u, v, k;
while(~scanf("%d", &n)){
memset(head, -, sizeof(head));
memset(Three_Max, , sizeof(Three_Max));
memset(cnt, , sizeof(cnt));
k = ;
for(int i = ;i < n;i ++){
if( == i){
scanf("%d", &cc);
val[] = cc;
}else{
scanf("%d%d", &u, &cc);
val[i] = cc;
AddEdge(u, i, k);
k ++;
}
}
dfs();
scanf("%d", &m);
for(int i = ;i < m;i ++){
scanf("%d", &u);
if(cnt[u] < ) printf("-1\n");
else{
for(int j = ;j < ;j ++) printf("%d ", Three_Max[u][j]);
printf("%d\n", Three_Max[u][]);
}
}
}
return ;
}

2.

 #include<iostream>
#include<climits>
#include<algorithm>
#include<cstring>
#include<cstdio>
#define MAX 11111
using namespace std;
int Three_Max[MAX][], val[MAX], cnt[MAX];
typedef struct{
int to, next;
}Node;
Node edge[MAX];
int head[MAX];
void AddEdge(int u, int v, int i){
edge[i].to = v;
edge[i].next = head[u];
head[u] = i;
}
bool cmp(int a, int b){
return a > b;
}
int *dfs(int id){
cnt[id] = ;
Three_Max[id][] = val[id];
for(int i = head[id];i != -;i = edge[i].next){
int u = edge[i].to;
int *temp = dfs(u);
for(int j = ;j <= ;j ++) Three_Max[id][j] = temp[j-];
sort(Three_Max[id] + , Three_Max[id]+, cmp);
cnt[id] += cnt[u];
}
return Three_Max[id];
}
int main(){
int n, m, cc, u, v, k;
while(~scanf("%d", &n)){
memset(head, -, sizeof(head));
memset(Three_Max, , sizeof(Three_Max));
memset(cnt, , sizeof(cnt));
k = ;
for(int i = ;i < n;i ++){
if( == i){
scanf("%d", &cc);
val[] = cc;
}else{
scanf("%d%d", &u, &cc);
val[i] = cc;
AddEdge(u, i, k);
k ++;
}
}
dfs();
scanf("%d", &m);
for(int i = ;i < m;i ++){
scanf("%d", &u);
if(cnt[u] < ) printf("-1\n");
else{
for(int j = ;j < ;j ++) printf("%d ", Three_Max[u][j]);
printf("%d\n", Three_Max[u][]);
}
}
}
return ;
}

ZOJ --- 3516 Tree of Three的更多相关文章

  1. ZOJ 3201 Tree of Tree

    树形DP.... Tree of Tree Time Limit: 1 Second      Memory Limit: 32768 KB You're given a tree with weig ...

  2. 【HDU】3516 Tree Construction

    http://acm.hdu.edu.cn/showproblem.php?pid=3516 题意:平面n个点且满足xi<xj, yi>yj, i<j.xi,yi均为整数.求一棵树边 ...

  3. HDOJ 3516 Tree Construction

    四边形优化DP Tree Construction Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Jav ...

  4. HDOJ 3516 Tree Construction 四边形优化dp

    原题链接:http://acm.hdu.edu.cn/showproblem.php?pid=3516 题意: 大概就是给你个下凸包的左侧,然后让你用平行于坐标轴的线段构造一棵树,并且这棵树的总曼哈顿 ...

  5. HDU 3516 Tree Construction (四边形不等式)

    题意:给定一些点(xi,yi)(xj,yj)满足:i<j,xi<xj,yi>yj.用下面的连起来,使得所有边的长度最小? 思路:考虑用区间表示,f[i][j]表示将i到j的点连起来的 ...

  6. HDU.3516.Tree Construction(DP 四边形不等式)

    题目链接 贴个教程: 四边形不等式学习笔记 \(Description\) 给出平面上的\(n\)个点,满足\(X_i\)严格单增,\(Y_i\)严格单减.以\(x\)轴和\(y\)轴正方向作边,使这 ...

  7. HDU 3516 Tree Construction

    区间$dp$,四边形优化. #pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio&g ...

  8. ZOJ - 3201 Tree of Tree (树形背包)

    题意:有一棵树,树上每个结点都有一个权值,求恰好包含k个结点的子树的最大权值. 设dp[i][j]为以结点i为根的树中包含j个结点的子树的最大权值,则可以把这个结点下的每棵子树中所包含的所有子树的大小 ...

  9. 【转载】ACM总结——dp专辑

    感谢博主——      http://blog.csdn.net/cc_again?viewmode=list       ----------  Accagain  2014年5月15日 动态规划一 ...

随机推荐

  1. WKWebView无法(通过URL schemes)跳转到其他App

    Custom scheme URL 在WKWebView中默认是不支持的 (但Safari可以). 我们可以通过NSError来进行一些处理从而使得程序可以正常跳转: func webView(web ...

  2. c语言数组不同初始化方式的结果

    第一种初始化方式: #include <stdio.h> int main() { int numbers[5]={12,14}; for (int i=0; i<5; i++) { ...

  3. ArcGis for WPF(1)

    这篇文章主要是讲窗体中怎么加载一张在线地图. 第一步:首先引用ESRI.ArcGIS.Client.dll类库. 第二步:在XAML中添加如下代码: <Window x:Class=" ...

  4. thinkphp 行为扩展

    网站程序在运行的过程每个过程都可以看做是一种行为,例如:运行应用,加载类,执行方法,加载模板,解析模板等,也就是说,我们在程序执行过程中每个 步骤都可以 定义一些点,我们可以在运行 程序的时候 检查 ...

  5. Windwos平台上ffmpeg解码音频并且保存到wav文件中

    先附上代码,测试通过 #include <stdio.h> #include <math.h> #include "libavutil/avstring.h" ...

  6. 暑假集训(1)第一弹 -----士兵队列训练问题(Hdu1276)

    Description 某部队进行新兵队列训练,将新兵从一开始按顺序依次编号,并排成一行横队,训练的规则如下:从头开始一至二报数,凡报到二的出列,剩下的向小序号方向靠拢,再从头开始进行一至三报数,凡报 ...

  7. Cron运行原理

    from:http://blog.chinaunix.net/uid-20682147-id-4977039.html 目录 目录 1 1. 前言 1 2. 示例 1 3. 工作过程 2 4. 一个诡 ...

  8. [python] virtualenv下解决matplotlib中文乱码

    1. 安装中文字体 一般系统自带wqy-microhei,其ttc文件位于/usr/share/fonts/truetype/wqy/wqy-microhei.ttc 2. 将ttc文件复制到pyth ...

  9. markdown与textile之间互相转换

    markdown与textile之间互相转换 redmine中默认使用的是textile那么从别的地方复制过来的markdown格式的内容需要进行转换 找到一款工具叫做pandoc http://jo ...

  10. 理解依赖注入(IOC)和学习Unity

    资料1: IOC:英文全称:Inversion of Control,中文名称:控制反转,它还有个名字叫依赖注入(Dependency Injection). 作用:将各层的对象以松耦合的方式组织在一 ...