解题报告 题意: n个插头m个设备k种转换器.求有多少设备无法插入. 思路: 定义源点和汇点,源点和设备相连,容量为1. 汇点和插头相连,容量也为1. 插头和设备相连,容量也为1. 可转换插头相连,容量也为inf(由于插头有无限个) #include <map> #include <queue> #include <cstdio> #include <vector> #include <cstring> #include <iostream…

解题报告 思路: 要求抢劫银行的伙伴(想了N多名词来形容,强盗,贼匪,小偷,sad.都认为不合适)不在同一个路口相碰面,能够把点拆成两个点,一个入点.一个出点. 再设计源点s连向银行位置.再矩阵外围套上一圈.连向汇点t 矩阵内的点,出点和周围的点的出点相连. #include <iostream> #include <cstring> #include <cstdio> #include <queue> #define M 500000 #define N…

C - A Plug for UNIX    You are in charge of setting up the press room for the inaugural meeting of the United Nations Internet eXecutive (UNIX), which has an international mandate to make the free flow of information and ideas on the Internet as cumb…

C - A Plug for UNIXTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=88038#problem/C Description You are in charge of setting up the press room for the inaugural meeting of the United Nations Internet…

  A Plug for UNIX  You are in charge of setting up the press room for the inaugural meeting of the United Nations Internet eXecutive (UNIX), which has an international mandate to make the free flow of information and ideas on the Internet as cumberso…

A Plug for UNIX Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13855   Accepted: 4635 Description You are in charge of setting up the press room for the inaugural meeting of the United Nations Internet eXecutive (UNIX), which has an int…

题面 You are in charge of setting up the press room for the inaugural meeting of the United Nations Internet eXecutive (UNIX), which has an international mandate to make the free flow of information and ideas on the Internet as cumbersome and bureaucra…

题意:给定n个插座,m个插头,k个转换器(x,y),转换器可以让插头x转成插头y.问最少有多少个插头被剩下. 题解: 最大流或者二分图匹配.然而我不知道怎么打二分图匹配..打了最大流.这题字符串比较坑爹,我就先把所有字符串编号(去重),然后给每个点编两个号,一个代表它作为插头的编号,一个代表它作为插座的编号.最坑的是uvaWA了好久最后发现结尾要有一个回车..... 还有数据范围也是个坑点,点起码要开500. 代码如下:(poj上是单组数据,uva和hdu都是多组) #include<cstdi…

网络流,关键在建图 建图思路在代码里 /* 最大流SAP 邻接表 思路:基本源于FF方法,给每个顶点设定层次标号,和允许弧. 优化: 1.当前弧优化(重要). 1.每找到以条增广路回退到断点(常数优化). 2.层次出现断层,无法得到新流(重要). 时间复杂度(m*n^2) */ #include <iostream> #include <cstdio> #include <cstring> #include <map> #define ms(a,b) mem…

最大流. 流可以对应一种分配方式. 显然最大流就可以表示最多匹配数 #include<cstdio> #include<algorithm> #include<cstring> using namespace std; + ; + ; ; const int inf = 0x3f3f3f3f; char s[maxn][maxl],t[maxl],t2[maxl]; int g[maxn],v[maxm],nex[maxm],f[maxm],eid; int id[ma…