hdu 5438 Ponds 拓扑排序

Ponds Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/contests/contest_showproblem.php?pid=1001&cid=621 Description Betty owns a lot of ponds, some of them are connected with other ponds by pipes, and there will not be more than one…

hdu 5438(类似拓扑排序)

Ponds Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 3178 Accepted Submission(s): 988 Problem Description Betty owns a lot of ponds, some of them are connected with other ponds by pipes, a…

HDU.2647 Reward(拓扑排序 TopSort)

HDU.2647 Reward(拓扑排序 TopSort) 题意分析裸的拓扑排序详解请移步算法学习拓扑排序(TopSort) 这道题有一点变化是要求计算最后的金钱数.最少金钱值是888,最少的差额是1,如1号的人比2号钱要多,2号至少是1号人的钱数+1.根据拓扑排序的思想,我们只需要类比二叉树的层序遍历,每次取出入度为0的节点做处理,累加金钱数量,然后再取出入度为0的节点处理.直到取出所有的点或者判断成环即可. 代码总览 #include <iostream> #include <…

HDU - 5438 Ponds（拓扑排序删点+并查集判断连通分量）

题目: 给出一个无向图,将图中度数小于等于1的点删掉,并删掉与他相连的点,直到不能在删为止,然后判断图中的各个连通分量,如果这个连通分量里边的点的个数是奇数,就把这些点的权值求和. 思路: 先用拓扑排序删点并更新各个点的度数,然后用并查集判断各个连通分量里边的点个数的奇偶性就ok了. 代码: #include <bits/stdc++.h> #include <cstdio> #include <cstring> #include <iostream> #i…

hdu 5438 Ponds（长春网络赛拓扑+bfs）

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5438 Ponds Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 2237 Accepted Submission(s): 707 Problem Description Betty owns a lot of ponds, som…

ACM: hdu 2647 Reward -拓扑排序

hdu 2647 Reward Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Description Dandelion's uncle is a boss of a factory. As the spring festival is coming , he wants to distribute rewards to his workers. Now he has a trouble…

HDU 2647 Reward (拓扑排序)

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2647 题意是给你n点m条有向边,叶子点(出度为0)上的值为888,父亲点为888+1,依次计算... 让你求最小的值,但是中间出现有向环就不行了,输出-1. 拓扑排序队列实现,因为叶子是最小的,所以把边反向就可以求了. //拓扑队列实现 //就是先把入度为0的先入队,然后每次出队的时候把相邻的点的入度减1,要是入度为0则再入队,不断循环直到队列为空 //时间复杂度为O(N + E) //这题就是把边…

HDU5438:Ponds(拓扑排序)

Ponds Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 3204 Accepted Submission(s): 995 Problem Description Betty owns a lot of ponds, some of them are connected with other ponds by pipes, a…

HDU 5438 Ponds

Ponds Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 282 Accepted Submission(s): 86 Problem Description Betty owns a lot of ponds, some of them are connected with other ponds by pipes, and…

hdu 4857 逃生拓扑排序+PQ，剥层分析

pid=4857">hdu4857 逃生题目是求拓扑排序,但不是依照字典序最小输出,而是要使较小的数排在最前面. 一開始的错误思路:给每一个点确定一个优先级(该点所能到达的最小的点).然后用拓扑排序+优先对列正向处理,正向输出.这是错误的.例如以下例子: 1 5 4 5 2 4 3 2 1 3 1 正确的解法:是反向建边.点大的优先级高,用拓扑排序+优先队列,逆向输出序列就可以. 依据每对限制,可确定拓扑序列,但此时的拓扑序列可能有多个(没有之间关系的点的顺序不定).本题要求较小的点排到…

HDU 1285 经典拓扑排序入门题

确定比赛名次 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 37566 Accepted Submission(s): 14659 Problem Description 有N个比赛队(1<=N<=500),编号依次为1,2,3,....,N进行比赛,比赛结束后,裁判委员会要将所有参赛队伍从前往后依次排名,但现在裁判委员会不能直…

HDU 2647 逆向拓扑排序

令每一个员工都有一个自己的等级level[i] , 员工等级越高,那么工资越高,为了使发的钱尽可能少,所以每一级只增加一单位的钱输入a b表示a等级高于b,那么我们反向添加边,令b—>a那么in[a]++,再进行拓扑排序,每次都让边的终点的level值大于起始点,那么要用max取较大值 #include <iostream> #include <cstdio> #include <cstring> #include <queue> using nam…

HDU 4917 Permutation 拓扑排序的计数

题意: 一个有n个数的排列,给你一些位置上数字的大小关系.求合法的排列有多少种. 思路: 数字的大小关系可以看做是一条有向边,这样以每个位置当点,就可以把整个排列当做一张有向图.而且题目保证有解,所以只一张有向无环图.这样子,我们就可以把排列计数的问题转化为一个图的拓扑排序计数问题. 拓扑排序的做法可以参见ZJU1346 . 因为题目中点的数量比较多,所以无法直接用状压DP. 但是题目中的边数较少,所以不是联通的,而一个连通块的点不超过21个,而且不同连通块之间可以看做相互独立的.所以我们可以对…

HDU 5638 Toposort 拓扑排序优先队列

Toposort 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5638 Description There is a directed acyclic graph with n vertices and m edges. You are allowed to delete exact k edges in such way that the lexicographically minimal topological sort of the gr…

hdu 5438 Ponds dfs

Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others) Problem Description Betty owns a lot of ponds, some of them are connected with other ponds by pipes, and there will not be more than one pipe between two ponds. Ea…

hdu 5438（拓扑+bfs）

题意:建图,删掉所有连接点小于2的点,直到不能删为止,问最后剩余的联通块中,点的数量是奇数的联通块中的点的权值和. 思路:拓扑删点,bfs计算 #include <iostream> #include <cstdio> #include <cstring> #include <vector> #include <queue> using namespace std; typedef long long ll; vector<int>q…

hdu 5638 Toposort (拓扑排序+线段树）

Toposort Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 499 Accepted Submission(s): 203 Problem Description There is a directed acyclic graph with n vertices and m edges. You are allowed…

HDU 5438 Ponds （DFS，并查集）

题意:给定一个图,然后让你把边数为1的结点删除,然后求连通块结点数为奇的权值和. 析:这个题要注意,如果删除一些结点后,又形成了新的边数为1的结点,也应该要删除,这是坑,其他的,先用并查集判一下环,然后再找连通环. 代码如下: #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstrin…

hdu 2647 Reward(拓扑排序+反图)

题目链接:https://vjudge.net/contest/218427#problem/C 题目大意: 老板要给很多员工发奖金, 但是部分员工有个虚伪心态, 认为自己的奖金必须比某些人高才心理平衡: 但是老板很人道, 想满足所有人的要求, 并且很吝啬,想画的钱最少输入若干个关系 a b a c c b 意味着a 的工资必须比b的工资高同时a 的工资比c高: c的工资比b高当出现环的时候输出-1 #include<iostream> #include<cstring> #…

HDU 2647--Reward【拓扑排序】

Reward Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 5636 Accepted Submission(s): 1712 Problem Description Dandelion's uncle is a boss of a factory. As the spring festival is coming , he w…

hdu 4857 逃生拓扑排序+逆向建图

逃生 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Problem Description 糟糕的事情发生啦,现在大家都忙着逃命.但是逃命的通道很窄,大家只能排成一行. 现在有n个人,从1标号到n.同时有一些奇怪的约束条件,每个都形如:a必须在b之前.同时,社会是不平等的,这些人有的穷有的富.1号最富,2号第二富,以此类推.有钱人就贿赂负责人,所以他们有一些好处. 负责人现在…

hdu 4857 逆向拓扑排序+反向输出

/*一组测试实例 4 4 2 3 1 2 4 */ #include<stdio.h> #include<string.h> #include<queue> using namespace std; #define N 31000 struct node { int u,v,next; }bian[N*10]; int head[N],yong,indegree[N],n,f[N],len; void init() { memset(head,-1,sizeof(hea…

hdu 2647 Reward(拓扑排序+优先队列）

Problem Description Dandelion's uncle is a boss of a factory. As the spring festival is coming , he wants to distribute rewards to his workers. Now he has a trouble about how to distribute the rewards. The workers will compare their rewards ,and some…

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