题目连接 http://poj.org/problem?id=2367 Genealogical tree Description The system of Martians' blood relations is confusing enough. Actually, Martians bud when they want and where they want. They gather together in different groups, so that a Martian can…

Genealogical tree Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 3674   Accepted: 2445   Special Judge Description The system of Martians' blood relations is confusing enough. Actually, Martians bud when they want and where they want. T…

Genealogical tree Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 8003   Accepted: 5184   Special Judge Description The system of Martians' blood relations is confusing enough. Actually, Martians bud when they want and where they want. T…

Genealogical tree Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 7101 Accepted: 4585 Special Judge Description The system of Martians' blood relations is confusing enough. Actually, Martians bud when they want and where they want. They ga…

题目链接 http://poj.org/problem?id=2367 题意就是给定一系列关系,按这些关系拓扑排序. #include<cstdio> #include<cstring> #include<queue> #include<vector> #include<algorithm> using namespace std; ; int ans; int n; int in[maxn]; //记录入度 int num[maxn]; //记…

火星人的血缘关系很奇怪,一个人可以有很多父亲,当然一个人也可以有很多孩子.有些时候分不清辈分会产生一些尴尬.所以写个程序来让n个人排序,长辈排在晚辈前面. 输入:N 代表n个人 1~n 接下来n行 第i行表示第i个人的孩纸,无序排列,可能为空.0代表一行输入结束. (大概我的智商真的不合适,否则怎么这么久了连个拓扑排序都写不好,T了三次..) 代码: /******************************************** Problem: 2367 User: Memory:…

题目:火星人的血缘关系,简单拓扑排序.很久没用邻接表了,这里复习一下. import java.util.Scanner; class edge { int val; edge next; } public class Main { static int n; static int MAXV = 1001; static edge head[] = new edge[MAXV]; static int in[]; static boolean vis[]; public static void…

一条标准的拓扑题解. 我这里的做法就是: 保存单亲节点作为邻接表的邻接点,这样就非常方便能够查找到那些点是没有单亲的节点,那么就能够输出该节点了. 详细实现的方法有非常多种的,比方记录每一个节点的入度,输出一个节点之后,把这个节点对于其它节点的入度去掉,然后继续查找入度为零的点输出.这个是一般的做法了,效果和我的程序一样的. 有兴趣的也能够參考下我这样的做法. #include <stdio.h> #include <string.h> #include <vector>…

题意:大概意思是--有一个家族聚集在一起,现在由家族里面的人讲话,辈分高的人先讲话.现在给出n,然后再给出n行数 第i行输入的数表示的意思是第i行的子孙是哪些数,然后这些数排在i的后面. 比如样例 5 0 4 5 1 0 1 0 5 3 0 3 0 1后面没有数 2后面有4 5 1 3后面有1 4后面有5 3 5后面有3 拓扑排序的一点小体会 (1)先把图储存下来,然后储存相应顶点的度数 (2)在图中选择没有前驱的点(即入度为0的点),加入队列中 (3)将以这一点为起点的边删去(将这一点指向的点…

Genealogical tree Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7285   Accepted: 4704   Special Judge Description The system of Martians' blood relations is confusing enough. Actually, Martians bud when they want and where they want. T…

Genealogical tree Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6025   Accepted: 3969   Special Judge Description The system of Martians' blood relations is confusing enough. Actually, Martians bud when they want and where they want. T…

Genealogical tree Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 2738 Accepted: 1838 Special Judge Description The system of Martians' blood relations is confusing enough. Actually, Martians bud when they want and where they want. They ga…

Genealogical tree Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 2920 Accepted: 1962 Special Judge Description The system of Martians' blood relations is confusing enough. Actually, Martians bud when they want and where they want. They ga…

Genealogical Tree Time limit: 1.0 secondMemory limit: 64 MB Background The system of Martians’ blood relations is confusing enough. Actually, Martians bud when they want and where they want. They gather together in different groups, so that a Martian…

Genealogical tree Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other) Total Submission(s) : 5   Accepted Submission(s) : 3 Special Judge Problem Description The system of Martians' blood relations is confusing enough. Ac…

POJ.3321 Apple Tree ( DFS序 线段树 单点更新 区间求和) 题意分析 卡卡屋前有一株苹果树,每年秋天,树上长了许多苹果.卡卡很喜欢苹果.树上有N个节点,卡卡给他们编号1到N,根的编号永远是1.每个节点上最多结一个苹果.卡卡想要了解某一个子树上一共结了多少苹果. 现在的问题是不断会有新的苹果长出来,卡卡也随时可能摘掉一个苹果吃掉.你能帮助卡卡吗? 前缀技能 边表存储树 DFS时间戳 线段树 首先利用边表将树存储下来,然后DFS打上时间戳.打上时间戳之后,我们就知道书上节点对…

[POJ2367]Genealogical tree Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5696   Accepted: 3729   Special Judge Description The system of Martians' blood relations is confusing enough. Actually, Martians bud when they want and where the…

Genealogical tree Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 6032 Accepted: 3973 Special Judge Description The system of Martians' blood relations is confusing enough. Actually, Martians bud when they want and where they want. They gat…

Genealogical tree Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 3658   Accepted: 2433   Special Judge Description The system of Martians' blood relations is confusing enough. Actually, Martians bud when they want and where they want. T…

id=10486" target="_blank" style="color:blue; text-decoration:none">POJ - 3321 Apple Tree Time Limit: 2000MS   Memory Limit: 65536KB   64bit IO Format: %I64d & %I64u Submit Status Description There is an apple tree outside o…

好抽象的树形DP......... Apple Tree Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 6411 Accepted: 2097 Description Wshxzt is a lovely girl. She likes apple very much. One day HX takes her to an apple tree. There are N nodes in the tree. Each nod…

http://poj.org/problem?id=2367 题意:给你n个数,从第一个数到第n个数,每一行的数字代表排在这个行数的后面的数字,直到0. 这是一个特别裸的拓扑排序的一个题目,拓扑排序我也是刚刚才接触,想法还是挺简单的.实现起来也不复杂. #include <stdio.h> #include <string.h> ],n; ][]; int topsort() { ;i<=n;i++) ;j<=n;j++) //遍历n次,找到第一个度为0的点,度为0的点…

Apple Tree Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 18623   Accepted: 5629 Description There is an apple tree outside of kaka's house. Every autumn, a lot of apples will grow in the tree. Kaka likes apple very much, so he has been…

Apple Tree Time Limit: 2000MS   Memory Limit: 65536K       Description There is an apple tree outside of kaka's house. Every autumn, a lot of apples will grow in the tree. Kaka likes apple very much, so he has been carefully nurturing the big apple t…

Apple Tree Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 25904   Accepted: 7682 Description There is an apple tree outside of kaka's house. Every autumn, a lot of apples will grow in the tree. Kaka likes apple very much, so he has been…

http://poj.org/problem?id=3237 题意:树链剖分.操作有三种:改变一条边的边权,将 a 到 b 的每条边的边权都翻转(即 w[i] = -w[i]),询问 a 到 b 的最大边权. 思路:一开始没有用区间更新,每次翻转的时候都更新到叶子节点,居然也能过,后来看别人的发现也是可以区间更新的. 第一种:无区间更新水过 #include <cstdio> #include <algorithm> #include <iostream> #inclu…

题目链接:http://poj.org/problem?id=2367 题意: 知道一个数n, 然后n行,编号1到n, 每行输入几个数,该行的编号排在这几个数前面,输出一种符合要求的编号名次排序. 拓扑排序: 先找入度为0的点,再根据这个点删掉与之相连的点之间的弧,入度减一. #include <stdio.h> ][]; ]; int main() { int n; scanf("%d",&n); ;i<=n;i++) { int to; while(sca…

题目:http://poj.org/problem?id=3321 动态更新某个元素,并且求和,显然是二叉索引树,但是节点的标号不连续,二叉索引树必须是连续的,所以需要转化成连续的,多叉树的形状已经建好,只要重新标号成连续的就行了. 感觉重新标号是这个题最难的地方,否则就是个纯水题了... 重新标号是看的别人的...用dfs遍历多叉树标号. #include <stdio.h> #include <stdlib.h> #include <string.h> ; ], n…

Tree   Description Give a tree with n vertices,each edge has a length(positive integer less than 1001). Define dist(u,v)=The min distance between node u and v. Give an integer k,for every pair (u,v) of vertices is called valid if and only if dist(u,v…

http://poj.org/problem?id=2486 典型的回溯题目:特别是状态方程用三维的来标记是否要走回路. 题意:一颗树,n个点(1-n),n-1条边,每个点上有一个权值,求从1出发,走V步,最多能遍历到的权值 思路: 树形dp,比较经典的一个树形dp.首先很容易就可以想到用dp[root][k]表示以root为根的子树中最多走k时所能获得的最多苹果数,接下去我们很习惯地会想到将k步在root的所有子结点中分配,也就是进行一次背包,就可以得出此时状态的最优解了,但是这里还有一个问题…