poj2942 双联通分量+交叉染色法判断二分图

时间：2016-05-03 01:55:03 阅读：199 评论：0 收藏：0 [点我收藏+]

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Knights of the Round Table

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 11805		Accepted: 3870

Description

Being a knight is a very attractive career: searching for the Holy Grail, saving damsels in distress, and drinking with the other knights are fun things to do. Therefore, it is not very surprising that in recent years the kingdom of King Arthur has experienced 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

The input contains several blocks of test cases. Each case begins with a line containing two integers 1 ≤ n ≤ 1000 and 1 ≤ m ≤ 1000000 . The number n is the number of knights. The next m lines describe which knight hates which knight. Each of these m lines 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

For each test case you have to output a single integer on a separate line: the number of knights that have to be expelled.

Sample Input

Sample Output

五一回家前开始想，今天回学校算是整个搞懂了吧。
题意：要求存在几个点能构成奇数环，则这几个点可以留下，判断需要淘汰几个点
思路：题目给的是互相之间的讨厌，所以第一步先构造like的反图
targan判断双连通分量(环)
判断这个环是否是奇环(交叉染色法)
这里还有一个前提定理，只要双连通分量的一个点是奇数环，其他点也会存在奇数环将其包围，画一下就明白了
还是非常有考验的题目啦

/*
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 888    d8P    d8b                                      d8888             888      888
 888   d8P     Y8P                                     d88888             888      888
 888  d8P                                             d88P888             888      888
 888d88K       888   88888b.     .d88b.              d88P 888   888d888   888888   88888b.    888  888   888d888
 8888888b      888   888 "88b   d88P"88b            d88P  888   888P"     888      888 "88b   888  888   888P"
 888  Y88b     888   888  888   888  888           d88P   888   888       888      888  888   888  888   888
 888   Y88b    888   888  888   Y88b 888          d8888888888   888       Y88b.    888  888   Y88b 888   888
 888    Y88b   888   888  888    "Y88888         d88P     888   888        "Y888   888  888    "Y88888   888
                                     888
                                Y8b d88P
                                 "Y88P"
 .d8888b.               888
d88P  Y88b              888
Y88b.                   888
 "Y888b.      8888b.    88888b.     .d88b.    888d888
    "Y88b.       "88b   888 "88b   d8P  Y8b   888P"
      "888   .d888888   888  888   88888888   888
Y88b  d88P   888  888   888 d88P   Y8b.       888
 "Y8888P"    "Y888888   88888P"     "Y8888    888
*/


#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <stack>
#include <queue>
#include <string>
#include <vector>

const int inf = 0x3f3f3f;
const int MAXN = 1e3+10;
const int MMAXN = 1e6+10;
struct edge{
    int next;
    int st;
    int to;
    int vis;
};
using namespace std;
int n,m;

stack<int>s;
int tG[MAXN][MAXN];
edge e[MMAXN];
int top;
int first[MAXN];
int dfn[MAXN];
int low[MAXN];
int mark[MAXN];
int col[MAXN];
int odd[MAXN];

void init(){
    top = 0;
    memset(tG,0,sizeof(tG));
    memset(dfn,0,sizeof(dfn));
    memset(low,0,sizeof(low));
    memset(odd,0,sizeof(odd));
    memset(first,-1,sizeof(first));
    while(!s.empty()){
        s.pop();
    }

}

void addege(int u,int v){
    e[top].st = u;
    e[top].to = v;
    e[top].next = first[u];
    e[top].vis = 0;
    first[u] = top++;
}


//交叉染色法判断是否为奇环
int find(int u){
    int v;

    for(int i=first[u];i!=-1;i=e[i].next){
        v = e[i].to;
        if(mark[v]){
            if(col[v]==-1){
                col[v] = !col[u];
                return find(v);
            }
            else if(col[v]==col[u])
                return 1;
        }
    }
    return 0;
}

void color(int u){
    int tmp;
    memset(mark,0,sizeof(mark));
    tmp = s.top();
    while(e[tmp].st!=u){
        mark[e[tmp].to] = 1;
        mark[e[tmp].st] = 1;
        s.pop();
        tmp = s.top();
    }
    mark[e[tmp].st] = 1;
    mark[e[tmp].to] = 1;
    s.pop();
    memset(col,-1,sizeof(col));
    col[u] = 1;
    if(find(u)){
        for(int i=1;i<=n;i++){
            if(mark[i]){
                odd[i] = 1;
            }
        }
    }
}

void dfs(int u,int step){
    dfn[u] = low[u] = step;
    int v;
    for(int i=first[u];i!=-1;i=e[i].next){
        if(e[i].vis)continue;
        e[i].vis = e[i^1].vis = 1;
        s.push(i);
        v = e[i].to;
        if(!dfn[v]){
            dfs(v,step+1);
            low[u] = min(low[u],low[v]);
            if(low[v]>=dfn[u])color(u); //每次不是双连通块了就要及时更新
        }
        else
            low[u] = min(low[u],dfn[v]);
    }

}

int main()
{
    int a,b;
    while(scanf("%d%d",&n,&m),n){
        int ans=0;
        init();
        
        for(int i=0;i<m;i++){
            scanf("%d%d",&a,&b);
            tG[a][b] = 1;
        }
        //构造反图
        for(int i=1;i<n;i++){
            for(int j=i+1;j<=n;j++){
                if(!tG[i][j]){
                    addege(i,j);
                    addege(j,i);
                }
            }
        }

        //寻找双连通块(环)
        for(int i=1;i<=n;i++){
            if(!dfn[i]){
                dfs(i,1);
            }
        }
        for(int i=1;i<=n;i++){
            if(!odd[i]){
                ans++;
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}

View Code

poj2942 双联通分量+交叉染色法判断二分图

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原文地址：http://www.cnblogs.com/EdsonLin/p/5453632.html

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