Legal or Not Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 11382    Accepted Submission(s): 5346 Problem Description ACM-DIY is a large QQ group where many excellent acmers get together. It is…

Problem Description ACM-DIY is a large QQ group where many excellent acmers get together. It is so harmonious that just like a big family. Every day,many "holy cows" like HH, hh, AC, ZT, lcc, BF, Qinz and so on chat on-line to exchange their ide…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5154 题解: 有向图判环. 1.用dfs,正在访问的节点标记为-1,已经访问过的节点标记为1,没有访问过的节点标记为0,如果访问到-1的节点说明说有环. #include<iostream> #include<cstdio> #include<cstring> #include<cmath> using namespace std; typedef long…

Dwarves 时间限制: 1 Sec  内存限制: 64 MB提交: 14  解决: 4[提交][状态][讨论版] 题目描述 Once upon a time, there arose a huge discussion among the dwarves in Dwarfland. The government wanted to introduce an identity card for all inhabitants. Most dwarves accept to be small, b…

传送门:http://oj.cnuschool.org.cn/oj/home/problem.htm?problemID=1042 试题描述: LZJ有一个问题想问问大家.他在写函数时有时候很头疼,如他写了这样几个函数: void f1(){   f2();   f3();}void f2(){   f3();}void f3(){   f1();}LZJ发现他无论怎么调换函数的位置,编译器总是不能通过编译,因为编译器规定调用的函数必须在当前函数之前写.还有一种情况是这样的:void f4(){…

HDU.3342 Legal or Not (拓扑排序 TopSort) 题意分析 裸的拓扑排序 根据是否成环来判断是否合法 详解请移步 算法学习 拓扑排序(TopSort) 代码总览 #include <iostream> #include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <sstream> #include…

Sorting It All Out Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 39602   Accepted: 13944 Description An ascending sorted sequence of distinct values is one in which some form of a less-than operator is used to order the elements from s…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3342 Legal or Not Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 7416    Accepted Submission(s): 3541 Problem Description ACM-DIY is a large QQ gr…

<题目链接> 题目大意: 给你 0~n-1 这n个点,然后给出m个关系 ,u,v代表u->v的单向边,问你这m个关系中是否产生冲突. 解题分析: 不难发现,题目就是叫我们判断图中是否存在环,存在环,则说明冲突.所以我们对图进行拓扑排序,如果该图中所有的点均能在拓扑排序中成为入度为0的点,则说明不含环(因为环中的点不可能入度为0). #include <cstdio> #include <cstring> #define rep(i,s,t) for(int i=s…

题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=3342 Legal or Not Description ACM-DIY is a large QQ group where many excellent acmers get together. It is so harmonious that just like a big family. Every day,many "holy cows" like HH, hh, AC, ZT, l…

题意: 一种游戏,2个人轮流控制棋子在一块有向图上移动,每次移动一条边,不能移动的人为输,无限循环则为平局,棋子初始位置为$S$ 现在有一个人可以同时控制两个玩家,问是否能使得第一个人必胜,并输出一个解,否则判断是否能平局 题解: 看到这个题首先我想到了强连通分量,但是事实证明求出强连通分量,缩点对解决问题没有什么帮助.... 能写一些看似正确的算法,但其实是假算法来的.. ........... 所以应该先分析策略,肯定是能赢就赢,不能赢就求平局,最后才算输 平局很好判断,有向图上,从$S$点…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3342 题目大意:n个点,m条有向边,让你判断是否有环. 解题思路:裸题,用dfs版的拓扑排序直接套用即可. 代码: #include<iostream> #include<cstdio> #include<cstring> using namespace std; ; int n,m; int vis[N]; bool G[N][N]; bool dfs(int u){ v…

Legal or Not Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 9125    Accepted Submission(s): 4231 Problem Description ACM-DIY is a large QQ group where many excellent acmers get together. It is…

Legal or Not Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6580    Accepted Submission(s): 3088 Problem Description ACM-DIY is a large QQ group where many excellent acmers get together. It is…

Legal or Not Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 5906    Accepted Submission(s): 2734 Problem Description ACM-DIY is a large QQ group where many excellent acmers get together. It is…

题意: 给出n个人的师徒关系,如有 a是b的师傅,b是c的师傅,c是a的师傅,这样则不合法,输出NO,否则输出YES. 思路: 每段关系可以看成一条有向边,从师傅指向徒弟,那么徒弟的徒子徒孙都不可能再指向其师傅或自己,所以不可能构成环. 两种方法 : 1,用拓扑的方法,每次去掉一个入度为0的点,全部点都去掉就是合法的.(下面代码用此法) 2,用深搜,记录路径,当搜到走过的路径上的点就是有环.每个点都可能重复搜多次,因为这不是树. #include <bits/stdc++.h> using n…

http://acm.hdu.edu.cn/showproblem.php?pid=1317 题意: 给出一个有向图,每到达一个点,都会加上或减去一些能量,我们要做的就是判断从1出发是否能到达n.初始能量有100,行走的途中能量不能小于等于0. 思路: 首先我们用floyd来判断一下1和n之间是否有通路. 其次就是bellman_ford算法来判正环了. #include <iostream> #include <cstring> #include <algorithm>…

Legal or Not Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other) Total Submission(s) : 42   Accepted Submission(s) : 32 Font: Times New Roman | Verdana | Georgia Font Size: ← → Problem Description ACM-DIY is a large QQ gr…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1269 题意:略 题解:trajan模版直接求强连通分量. #include <iostream> #include <cstring> #include <cstdio> using namespace std; const int N = 1e4 + 10; const int M = 1e5 + 10; struct Edge { int v , next; }edge…

Legal or NotTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 11170    Accepted Submission(s): 5230 Problem DescriptionACM-DIY is a large QQ group where many excellent acmers get together. It is s…

一道极其水的拓扑排序……但是我还是要把它发出来,原因很简单,连错12次…… 题意也很裸,前面的废话不用看,直接看输入 输入n, m表示从0到n-1共n个人,有m组关系 截下来m组,每组输入a, b表示a指向b,或者b指向a也行. 输入n == 0时结束 如果可以拓扑排序,输出"YES",否则输出"NO".每组输出占一行. 给两种代码吧,一种用邻接矩阵,另一种我也不知道叫什么好…… 邻接矩阵—— #include <cstdio> #include <…

重边这样的东西   仅仅能呵呵 就是裸裸的拓扑排序 假设恩可以排出来就YES . else  NO 仅仅须要所有搜一遍就好了 #include <cstdio> #include <cstring> int mapp[101][101]; int d[101]; int n,m; int a,b; int main() { while(scanf("%d%d",&n,&m)!=EOF) { if(n==0)break; memset(d,0,si…

题意:给出n,m,人的编号为 0到n-1,再给出m个关系,问能不能够进行拓扑排序 #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; int indegree[1005],queue[1005],head[1005]; struct Edge { int to; int next; } edge[1005]; int ma…

Matching In Multiplication Problem DescriptionIn the mathematical discipline of graph theory, a bipartite graph is a graph whose vertices can be divided into two disjoint sets U and V (that is, U and V are each independent sets) such that every edge…

DZY Loves Topological Sorting Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5195 Description A topological sort or topological ordering of a directed graph is a linear ordering of its vertices such that for ev…

Sorting It All Out Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 33184   Accepted: 11545 Description An ascending sorted sequence of distinct values is one in which some form of a less-than operator is used to order the elements from s…

Rank of Tetris Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5415    Accepted Submission(s): 1514 Problem Description 自从Lele开发了Rating系统,他的Tetris事业更是如虎添翼,不久他遍把这个游戏推向了全球. 为了更好的符合那些爱好者的喜好,Lele又想了…

逃生 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 7736    Accepted Submission(s): 2248 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…

Gym Class Time Limit: 6000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 768    Accepted Submission(s): 309 Problem Description 众所周知,度度熊喜欢各类体育活动. 今天,它终于当上了梦寐以求的体育课老师.第一次课上,它发现一个有趣的事情.在上课之前,所有同学要排成一列, 假设最开始每个人有…

Harry and Magical Computer Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description In reward of being yearly outstanding magic student, Harry gets a magical computer. When the computer begins to deal wi…