先上题目:

1213. Cockroaches!

Time limit: 1.0 second
Memory limit: 64 MB
  It's well-known that the most tenacious of life species on the Earth are cockroaches. They live everywhere if only there in food. And as far as they are unpretentious in food you can find them absolutely everywhere.
  A little Lyosha studies at school on a Space station. During one of the school competitions his class has reached the final. A task of the final contest is to exterminate all the cockroaches in the cargo module within minimal time.
  Within the long history of the competitions a unified tactics was worked out. The tactics is as follows: a poison gas is let in one of the module compartments and after that the baffle that separates the compartment from one of the adjacent ones is opened.  Cockroaches can't stand the smell of the gas and run to the other compartment. When there's no cockroaches in the treated compartment the baffle is closed. Afterwards analogously the next compartment is treated, and so on. The goal is to move all the cockroaches to the floodgate of the cargo module. Then the outward door is opened and all the cockroaches are engulfed by an open Space.
  Lyosha is responsible for programming the control board of the baffles in his team. The baffles are opened slowly, so it's very important to make do with minimal number of baffle openings in order to win in the contest. Your task is to help Lyosha to compute this number.

Input

  The first line contains a name of the floodgate compartment. Each of the next lines contains description of one of the baffles — the names of two compartments separated with a dash (-). The last line contains the only symbol "#". There are cockroaches in all the compartments of the module at first. It's possible to get to the floodgate from every compartment of the module passing several baffles. The total number of compartments doesn't exceed 30. The name of a compartment consists of no more than 20 Latin letters and digits. The large and the small letters should be distinguished.

Output

  Your program is to output the only number — the minimal amount of baffles that should be opened (and then closed) in order to move all the cockroaches to the floodgate.

Sample

input output
Gateway
Machinery-Gateway
Machinery-Control
Control-Central
Control-Engine
Central-Engine
Storage-Gateway
Storage-Waste
Central-Waste
#
6

  题意:给你一个终点的名字,给你一个连通图上不同点之间的关系,点上都有蟑螂。现在每一次可以选着一条边,将一个点上的蟑螂都赶到边的另一端,然后这条边会消失,问至少需要多少步操作才能将所有的蟑螂赶到终点。

  其实这就是在问一个联通图上面保留多少条边可以形成一棵树。树的结构是有n个点就有n-1条边。所以只需要记录每一个出现过的端点的名字各一次,树木减一就得出正确答案了。这里可以使用STL的set容器比较简单,当然用map也可以,不过set会快一点,也可以写一棵字典树,不过那样就很费时间了。

上代码:

 #include <cstdio>
#include <cstring>
#include <string>
#include <set> using namespace std; typedef struct{
char k[];
}st; set<string> s; int main()
{
st a,b,c;
string x,y;
int tot;
//freopen("data.txt","r",stdin);
scanf("%s",a.k);
x=a.k;
getchar();
tot=;
s.clear();
s.insert(x);
tot++;
while(scanf("%s",a.k),strcmp(a.k,"#")){
getchar();
int l=strlen(a.k);
int j=;
for(int i=;i<l;i++){
if(a.k[i]=='-'){
b.k[j]='\0';
strcpy(c.k,a.k+i+);
break;
}
b.k[j]=a.k[i];
j++;
}
x=b.k;
y=c.k;
if(s.count(x)<=){
s.insert(x);
tot++;
}
if(s.count(y)<=){
s.insert(y);
tot++;
}
}
printf("%d\n",tot-);
return ;
}

1213

Timus - 1213 - Cockroaches!的更多相关文章

  1. LightOj 1213 - Fantasy of a Summation(推公式 快速幂)

    题目链接:http://lightoj.com/volume_showproblem.php?problem=1213 #include <stdio.h> int cases, case ...

  2. codeforces 719B:Anatoly and Cockroaches

    Description Anatoly lives in the university dorm as many other students do. As you know, cockroaches ...

  3. hdu 1213 How Many Tables 解题报告

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1213 有关系(直接或间接均可)的人就坐在一张桌子,我们要统计的是最少需要的桌子数. 并查集的入门题,什 ...

  4. Timus OJ 1997 Those are not the droids you're looking for (二分匹配)

    题目链接:http://acm.timus.ru/problem.aspx?space=1&num=1997 这个星球上有两种人,一种进酒吧至少玩a小时,另一种进酒吧最多玩b小时. 下面n行是 ...

  5. 1213 How Many Tables(简单并查集)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1213 简单并查集,统计单独成树的数量. 代码: #include <stdio.h> #i ...

  6. Codeforces 719B Anatoly and Cockroaches

    B. Anatoly and Cockroaches time limit per test:1 second memory limit per test:256 megabytes input:st ...

  7. Timus Online Judge:ural:1006. Square Frames

    原题链接:http://acm.timus.ru/problem.aspx?space=1&num=1006 看到题第一反应:这玩意怎么读入…… 本地的话因为是全角字符,会占两个位置,所以需要 ...

  8. HDU 1213 How Many Tables(模板——并查集)

    题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1213 Problem Description Today is Ignatius' birthday ...

  9. Timus 1132 Square Root(二次剩余)

    http://acm.timus.ru/problem.aspx?space=1&num=1132 题意: 求 x^2 ≡ n mod p  p是质数 的 解 本题中n>=1 特判p=2 ...

随机推荐

  1. [POJ 1745] Divisbility

    [题目链接] http://poj.org/problem?id=1745 [算法] DP [代码] #include <algorithm> #include <bitset> ...

  2. 【BZOJ 1230】 开关灯

    [题目链接] https://www.lydsy.com/JudgeOnline/problem.php?id=1230 [算法] 线段树 [代码] #include<bits/stdc++.h ...

  3. 通过Ajax和SpringBoot交互的示例

    转自:https://blog.csdn.net/oppo5630/article/details/52093898/

  4. E20170906-mk

    portrait   n. 肖像,肖像画; 模型,标本; 半身雕塑像; 人物描写; orientation  n. 方向,定位,取向,排列方向; 任职培训; (外交等的) 方针[态度]的确定; 环境判 ...

  5. 洛谷P1402 酒店之王(二分图)

    P1402 酒店之王 题目描述 XX酒店的老板想成为酒店之王,本着这种希望,第一步要将酒店变得人性化.由于很多来住店的旅客有自己喜好的房间色调.阳光等,也有自己所爱的菜,但是该酒店只有p间房间,一天只 ...

  6. java 中接口的概念

    接口接口在java中是一个抽象的类型,是抽象方法的集合,接口通常使用interface来声明,一个类通过继承接口的方式从而继承接口的抽象方法.接口并不是类,编写接口的方式和类的很相似,但是他们属于不同 ...

  7. Windows:Word,PPT,EXCEL com+组件配置

    本文所涉及到配置前提: 服务器必须安装Office套件(Word,PPT,Excel) 第一部分 Word Com+组件权限配置 1.cmd模式输入dcomcnfg 2.找到Microsoft Wor ...

  8. Oracle11g聚合函数

    聚合函数就是基于多行数据返回一行结果,下面就是Oracle提供的一些列聚合函数: AVG COLLECT CORR CORR_* COUNT COVAR_POP COVAR_SAMP CUME_DIS ...

  9. 35个jquery小技巧

    1. 禁止右键点击 ? 1 2 3 4 5 $(document).ready(function(){     $(document).bind("contextmenu",fun ...

  10. ES6 arrow function

    语法: () => { … } // 零个参数用 () 表示: x => { … } // 一个参数可以省略 (): (x, y) => { … } // 多参数不能省略 (): 当 ...