http://poj.org/problem?id=2230 Description Bessie's been appointed the new watch-cow for the farm. Every night, it's her job to walk across the farm and make sure that no evildoers are doing any evil. She begins at the barn, makes her patrol, and the…

bzoj1123: [POI2008]BLO poj3694 先e-DCC缩点,此时图就变成了树,树上每一条边都是桥.对于添加边的操作,相当于和树上一条路径构环,导致该路径上所有边都不成为桥.那么找这条新加边的最近公共祖先,把路径上的所有没被删掉的桥的数量计算出来,未操作之前桥的个数减去该值就是当前答案.中间因为一条边会多次删除,没有意义,可以采取并查集路径压缩的思想,直接指向下一个没有被删的桥 #include<cstdio> #include<iostream> #includ…

题目链接:http://poj.org/problem?id=2230 题目: 题意:给你m条路径,求一条路径使得从1出发最后回到1,并满足每条路径都恰好被沿着正反两个方向经过一次. 思路:由于可以回到起点,并且题目保证有解,所以本题是欧拉回路.输出路径有两种方法,一种是递归实现,一种是用栈处理,不过两种速度差了1s,不知道是不是我递归没处理好~ 代码实现如下(第一种为栈输出路径,对应797ms,第二种为递归,对应1782ms): #include <set> #include <map…

Watchcow Time Limit: 3000MS Memory Limit: 65536K Total Submissions: 6172Accepted: 2663 Special Judge Description Bessie's been appointed the new watch-cow for the farm. Every night, it's her job to walk across the farm and make sure that no evildoers…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4392 题目大意: 一个图,要将每条边恰好遍历两遍,而且要以不同的方向,还要回到原点. dfs解法                   借鉴于->大佬博客 #include <iostream> #include <vector> #include <algorithm> using namespace std; #define arrsize 10001 int n,…

Watchcow Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 7473   Accepted: 3270   Special Judge Description Bessie's been appointed the new watch-cow for the farm. Every night, it's her job to walk across the farm and make sure that no ev…

Watchcow Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 7512   Accepted: 3290   Special Judge Description Bessie's been appointed the new watch-cow for the farm. Every night, it's her job to walk across the farm and make sure that no ev…

//网络流判定混合图欧拉回路 //通过网络流使得各点的出入度相同则possible,否则impossible //残留网络的权值为可改变方向的次数,即n个双向边则有n次 //Time:157Ms Memory:348K #include <iostream> #include<cstring> #include<cstdio> #include<algorithm> #include<queue> using namespace std; #de…

注意:找出一条欧拉回路,与判定这个图能不能一笔联通...是不同的概念 c++奇怪的编译规则...生不如死啊... string怎么用啊...cincout来救? 可以直接.length()我也是长见识了... CE怎么办啊...g++来救? #include<cstdio> #include<iostream> #include<cstring> #include<cmath> #include<algorithm> #define N 2020…

 FZU 2112 Tickets Time Limit:3000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Practice Description You have won a collection of tickets on luxury cruisers. Each ticket can be used only once, but can be used in either direction betwee…