poj 2942 Knights of the Round Table 圆桌骑士(双连通分量模板题)
Time Limit: 7000MS | Memory Limit: 65536K | |
Total Submissions: 9169 | Accepted: 2960 |
Description
an unprecedented increase in the number of knights. There are so many knights now, that it is very rare that every Knight of the Round Table can come at the same time to Camelot and sit around the round table; usually only a small group of the knights isthere,
while the rest are busy doing heroic deeds around the country.
Knights can easily get over-excited during discussions-especially after a couple of drinks. After some unfortunate accidents, King Arthur asked the famous wizard Merlin to make sure that in the future no fights break out between the knights. After studying
the problem carefully, Merlin realized that the fights can only be prevented if the knights are seated according to the following two rules:
- The knights should be seated such that two knights who hate each other should not be neighbors at the table. (Merlin has a list that says who hates whom.) The knights are sitting around a roundtable, thus every knight has exactly two neighbors.
- An odd number of knights should sit around the table. This ensures that if the knights cannot agree on something, then they can settle the issue by voting. (If the number of knights is even, then itcan happen that ``yes" and ``no" have the same number of
votes, and the argument goes on.)
Merlin will let the knights sit down only if these two rules are satisfied, otherwise he cancels the meeting. (If only one knight shows up, then the meeting is canceled as well, as one person cannot sit around a table.) Merlin realized that this means that
there can be knights who cannot be part of any seating arrangements that respect these rules, and these knights will never be able to sit at the Round Table (one such case is if a knight hates every other knight, but there are many other possible reasons).
If a knight cannot sit at the Round Table, then he cannot be a member of the Knights of the Round Table and must be expelled from the order. These knights have to be transferred to a less-prestigious order, such as the Knights of the Square Table, the Knights
of the Octagonal Table, or the Knights of the Banana-Shaped Table. To help Merlin, you have to write a program that will determine the number of knights that must be expelled.
Input
contains two integers k1 and k2 , which means that knight number k1 and knight number k2 hate each other (the numbers k1 and k2 are between 1 and n ).
The input is terminated by a block with n = m = 0 .
Output
Sample Input
5 5
1 4
1 5
2 5
3 4
4 5
0 0
Sample Output
2
Hint
搜索双连通分量。深度优先搜索过程中,用一个栈保存所有经过的节点,判断割点,碰到割点就标记当前栈顶的结点并退栈,直到当前结点停止并标记当前割点。标记过的结点处于同一个双连通分量。
交叉染色搜索奇圈。在一个节点大于2的双连通分量中,必定存在一个圈经过该连通分量的所有结点;如果这个圈是奇圈,则该连通分量内的所有的点都满足条件;若这个圈是偶圈,如果包含奇圈,则必定还有一个奇圈经过所有剩下的点。因此一个双连通分量中只要存在一个奇圈,那么该双联通分量内的所有的点都处于一个奇圈中,在题目中,即武士可以坐成一圈。根据这个性质,只需要在一个双联通分量内找奇圈即可判断该双联通分量是否满足条件。交叉染色法就是在dfs的过程中反复交换着用两种不同的颜色对点染色,若某次dfs中当前结点的子节点和当前结点同色,则找到奇圈。
参考于:http://www.cnblogs.com/wuyiqi/archive/2011/10/19/2217911.html
#include "stdio.h"
#include "string.h" #define N 1010 int time;
int n,m;
bool map[N][N]; struct node
{
int x,y;
//int weight;
bool visit; //用来标记该边是否已经访问
int next;
}edge[2*N*N];
int idx,head[N]; bool odd[N];
bool mark[N]; //标记点是否为当前双连通分量中的元素
int low[N],dfn[N];
int st[N*N],top; //模拟栈
int col[N]; void Read_date();
inline int MIN(int a,int b) { return a<b?a:b; }
void Init(){ idx=0; memset(head,-1,sizeof(head)); }
void Add(int x,int y)
{
edge[idx].x = x;
edge[idx].y = y;
edge[idx].visit = false;
edge[idx].next = head[x];
head[x] = idx++;
} bool find(int x) //判断当前双连通分量是否为二分图
{
int i,y;
for(i=head[x]; i!=-1; i=edge[i].next)
{
y = edge[i].y;
if(mark[y])
{
if(col[y]==-1)
{
col[y] = !col[x];
return find(y);
}
else if(col[y]==col[x])
return false; //不是二分图,返回false
}
}
return true; //是二分图,返回true
} void Color(int x)
{
int i;
memset(mark,false,sizeof(mark));
while(1)
{
i = st[top];
top--;
mark[edge[i].x] = true;
mark[edge[i].y] = true;
if(edge[i].x==x) break;
}
memset(col,-1,sizeof(col));
col[x] = 0;
if(!find(x)) //双连通分量不是二分图,则这些点全部可以
{
for(i=1; i<=n; ++i)
{
if(mark[i])
odd[i] = 1;
}
}
} void DFS(int x)
{
int i,y;
low[x] = dfn[x] = ++time;
for(i=head[x]; i!=-1; i=edge[i].next)
{
y = edge[i].y;
if(edge[i].visit) continue;
edge[i].visit = edge[i^1].visit = 1;
st[++top] = i;
if(!dfn[y])
{
DFS(y);
low[x] = MIN(low[x],low[y]);
if(low[y]>=dfn[x]) //找到割顶或者为根节点
Color(x);
}
else
low[x] = MIN(low[x],dfn[y]);
}
} int Solve()
{
int i;
int num=0;
time = 0;
top = 0;
memset(dfn,0,sizeof(dfn));
memset(odd,false,sizeof(odd));
for(i=1; i<=n; ++i)
{
if(!dfn[i]) //表示点i未被访问过
DFS(i); //以i为根节点找双连通分量
}
for(i=1; i<=n; ++i)
{
if(!odd[i])
num++;
}
return num;
} int main()
{
while(scanf("%d%d",&n,&m),n||m)
{
Read_date();
printf("%d\n",Solve());
}
return 0;
} void Read_date()
{
int i,j;
int x,y;
memset(map,true,sizeof(map));
while(m--)
{
scanf("%d %d",&x,&y);
map[x][y] = map[y][x] = false;
}
Init();
for(i=1; i<=n; ++i)
{
for(j=i+1; j<=n; ++j)
{
if(map[i][j])
{
Add(i,j);
Add(j,i);
}
}
}
}
poj 2942 Knights of the Round Table 圆桌骑士(双连通分量模板题)的更多相关文章
- POJ 2942 Knights of the Round Table (点双连通分量)
题意:多个骑士要开会,3人及以上才能凑一桌,其中部分人已经互相讨厌,肯定不坐在同一桌的相邻位置,而且一桌只能奇数个人才能开台.给出多个人的互相讨厌图,要求多少人开不成会(注:会议不要求同时进行,一个人 ...
- poj 2942 Knights of the Round Table(无向图的双连通分量+二分图判定)
#include<cstdio> #include<cstring> #include<cmath> #include<cstdlib> #includ ...
- POJ 2942 Knights of the Round Table 黑白着色+点双连通分量
题目来源:POJ 2942 Knights of the Round Table 题意:统计多个个骑士不能參加随意一场会议 每场会议必须至少三个人 排成一个圈 而且相邻的人不能有矛盾 题目给出若干个条 ...
- POJ2942 UVA1364 Knights of the Round Table 圆桌骑士
POJ2942 洛谷UVA1364(博主没有翻墙uva实在是太慢了) 以骑士为结点建立无向图,两个骑士间存在边表示两个骑士可以相邻(用邻接矩阵存图,初始化全为1,读入一对憎恨关系就删去一条边即可),则 ...
- poj 2942 Knights of the Round Table - Tarjan
Being a knight is a very attractive career: searching for the Holy Grail, saving damsels in distress ...
- POJ 2942 Knights of the Round Table
Knights of the Round Table Time Limit: 7000MS Memory Limit: 65536K Total Submissions: 10911 Acce ...
- POJ 2942 Knights of the Round Table - from lanshui_Yang
Description Being a knight is a very attractive career: searching for the Holy Grail, saving damsels ...
- 【LA3523】 Knights of the Round Table (点双连通分量+染色问题?)
Being a knight is a very attractive career: searching for the Holy Grail, saving damsels in distress ...
- poj 2942 Knights of the Round Table(点双连通分量+二分图判定)
题目链接:http://poj.org/problem?id=2942 题意:n个骑士要举行圆桌会议,但是有些骑士相互仇视,必须满足以下两个条件才能举行: (1)任何两个互相仇视的骑士不能相邻,每个骑 ...
随机推荐
- C#接口知识大全收藏
第一节 接口慨述 接口(interface)用来定义一种程序的协定.实现接口的类或者结构要与接口的定义严格一致.有了这个协定,就可以抛开编程语言的限制(理论上).接口可以从多个基接口继承,而类或结构可 ...
- C#中的lock关键字有何作用
作为C#的程序员来说,在遇到线程同步的需求时最常用的就是lock关键字.但如何正确并有效地使用lock,却是能否高效地达到同步要求的关键.正因为如此,程序员需要完全理解lock究竟为程序做了什么. 所 ...
- 重新想象 Windows 8 Store Apps (49) - 输入: 获取输入设备信息, 虚拟键盘, Tab 导航, Pointer, Tap, Drag, Drop
[源码下载] 重新想象 Windows 8 Store Apps (49) - 输入: 获取输入设备信息, 虚拟键盘, Tab 导航, Pointer, Tap, Drag, Drop 作者:weba ...
- Free Slideshow, Gallery And Lightboxes Scripts
http://bootstraphelpers.codeplex.com/SourceControl/list/changesets https://github.com/gordon-matt/Bo ...
- JMS学习(一)基本概念
这两天面试了一两个公司,由于简历中的最近一个项目用到了JMS,然而面试官似乎对这个很感兴趣,所以都被问到了,但可惜的是,我除了说我们使用了JMS外,面对他们提出的一些关于JMS的问题,我回答得相当差, ...
- mysql innodb表 utf8 gbk占用空间相同,毁三观
昨天因为发生字符集转换相关错误,今天想验证下utf8和gbk中英文下各自空间的差距.这一测试,绝对毁三观,无论中文还是中文+英文,gbk和utf8占用的实际物理大小完全相同,根本不是理论上所述的“UT ...
- C++之内联函数与constexpr
inline 函数 规模小,流程直接且频繁调用 cout<<shortString(s1,s2)<<endl; = cout<<(s1.size()<s2.s ...
- arcgis python 更新顺序号
i = 0def myFun(): global i i=i +1 return i myFun() ========================== accumulate( ) total = ...
- unity下载文件三(http异步下载)
异步下载,顾名思义就是不影响你主线程使用客户端的时候,人家在后台搞你的明堂. 直接入主题,既然要下载,首先得请求,请求成功之后进行回调,这就是一个异步过程,异步回调的时间不可控. 1.首先请求下载. ...
- Web应用程序系统的多用户权限控制设计及实现-数据库设计【2】
搭建一个Web权限配置的系统,需要以下五张数据表:人员表,分组表,页面表,目录表,操作权限表.各张数据表中用到的id均为自增1的标识,每张数据表的定义如下: 1.1人员表(operatorinfo)