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>…

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…

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…

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: 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: 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…

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…

裸拓扑排序. 拓扑排序 用一个队列实现,先把入度为0的点放入队列.然后考虑不断在图中删除队列中的点,每次删除一个点会产生一些新的入度为0的点.把这些点插入队列. 注意:有向无环图 g[] : g[i]表示从点i连出去的边 L[] :拓扑排序的结构 code: #include <cstdio> #include <cstring> #include <queue> using namespace std; const int maxn = 100 + 5; vector…

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: 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: 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…

[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…

今天网易的笔试,妹的,算法题没能A掉,虽然按照思路写了出来,但是尼玛好歹给个测试用例的格式呀,吐槽一下网易的笔试出的太烂了. 就一道算法题,比较石子重量,个人以为解法应该是拓扑排序. 就去POJ找了道拓扑排序的题:POJ2367 直接上代码吧: #include<stdio.h> #include<string> #define clr(x) memset(x,0,sizeof(x)) ][]; ]; ]; using namespace std; int main() { int…

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 : 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…

Genealogical tree poj-2367 题目大意:给你一个n个点关系网,求任意一个满足这个关系网的序列,使得前者是后者的上级. 注释:1<=n<=100. 想法:刚刚学习toposort,什么是toposort? 就是每一个点的遍历或选取有先决条件,那么我们可以通过队列或者栈将控制当前点的点先遍历,这样弹栈或出队的序列,就是toposort的结果 我们先附上模板 int v[100100];//入度 bool map[110][110]={false};//是否存在控制与被控制…

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…

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…

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…

地址 http://poj.org/problem?id=2431 题解 朴素想法就是dfs 经过该点的时候决定是否加油 中间加了一点剪枝 如果加油次数已经比已知最少的加油次数要大或者等于了 那么就剪枝 然而 还是TLE了 TLE代码 #include <iostream> #include <vector> #include <algorithm> #include <queue> using namespace std; vector<pair&l…

地址 http://poj.org/problem?id=1064 题解 二分即可 其实 对于输入与精度计算不是很在行 老是被卡精度 后来学习了一个函数 floor 向负无穷取整 才能ac 代码如下 #include <iostream> #include <vector> #include <math.h> #include <algorithm> using namespace std; vector<double> v; int n, k;…

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…

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 have one parent as well as ten. Nobody will be surpris…

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…

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…