亚洲免费在线-亚洲免费在线播放-亚洲免费在线观看-亚洲免费在线观看视频-亚洲免费在线看-亚洲免费在线视频

The Accomodation of Students

系統 1704 0

The Accomodation of Students

Time Limit: 5000/1000 MS (Java/Others)????Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1462????Accepted Submission(s): 716

?

Problem Description
There are a group of students. Some of them may know each other, while others don't. For example, A and B know each other, B and C know each other. But this may not imply that A and C know each other.

?

Now you are given all pairs of students who know each other. Your task is to divide the students into two groups so that any two students in the same group don't know each other.If this goal can be achieved, then arrange them into double rooms. Remember, only paris appearing in the previous given set can live in the same room, which means only known students can live in the same room.

?

Calculate the maximum number of pairs that can be arranged into these double rooms.
?

?

Input
For each data set:
The first line gives two integers, n and m(1<n<=200), indicating there are n students and m pairs of students who know each other. The next m lines give such pairs.

?

Proceed to the end of file.

?

?

?

Output
If these students cannot be divided into two groups, print "No". Otherwise, print the maximum number of pairs that can be arranged in those rooms.
?

?

Sample Input
4 4
1 2
1 3
1 4
2 3
6 5
1 2
1 3
1 4
2 5
3 6
?

?

Sample Output
No
3
?

?

Source
2008 Asia Harbin Regional Contest Online
?

?

Recommend
gaojie

又是一遍AC,雖然還是模板題,但是一個小時4個1a讓我今天下午很愉快啊有木有.

      
        #include<stdio.h>
        
           #include
        
        <
        
          string
        
        .h>


        
          int
        
        
           N,M; 
        
        
          int
        
         color[
        
          300
        
        ],match[
        
          300
        
        
          ]; 
        
        
          bool
        
         visit[
        
          300
        
        ],G[
        
          300
        
        ][
        
          300
        
        
          ],flag; 
        
        
          void
        
         draw(
        
          int
        
         k,
        
          int
        
        
           cc) { 
        
        
          if
        
         (color[k]!=-
        
          1
        
         && color[k]!=
        
          cc) { flag
        
        =
        
          false
        
        
          ; 
        
        
          return
        
        
          ; } 
        
        
          if
        
         (color[k]==cc) 
        
          return
        
        
          ; color[k]
        
        =
        
          cc; 
        
        
          int
        
         c=
        
          1
        
        -
        
          cc; 
        
        
          for
        
         (
        
          int
        
         i=
        
          1
        
        ;i<=N;i++
        
          ) 
        
        
          if
        
         (G[k][i] &&
        
           flag) draw(i,c); } 
        
        
          bool
        
         DFS(
        
          int
        
        
           k) { 
        
        
          int
        
        
           t; 
        
        
          for
        
         (
        
          int
        
         i=
        
          1
        
        ;i<=N;i++
        
          ) 
        
        
          if
        
         (G[k][i] && !
        
          visit[i]) { visit[i]
        
        =
        
          1
        
        
          ; t
        
        =
        
          match[i]; match[i]
        
        =
        
          k; 
        
        
          if
        
         (t==-
        
          1
        
         || DFS(t)) 
        
          return
        
        
          true
        
        
          ; match[i]
        
        =
        
          t; } 
        
        
          return
        
        
          false
        
        
          ; } 
        
        
          int
        
        
           Max_match() { 
        
        
          int
        
         ans=
        
          0
        
        
          ; memset(match,
        
        -
        
          1
        
        ,
        
          sizeof
        
        
          (match)); 
        
        
          for
        
         (
        
          int
        
         i=
        
          1
        
        ;i<=N;i++
        
          ) { memset(visit,
        
        
          0
        
        ,
        
          sizeof
        
        
          (visit)); 
        
        
          if
        
         (DFS(i)) ans++
        
          ; } 
        
        
          return
        
        
           ans; } 
        
        
          int
        
        
           main() { 
        
        
          while
        
         (scanf(
        
          "
        
        
          %d%d
        
        
          "
        
        ,&N,&M)!=
        
          EOF) { memset(G,
        
        
          0
        
        ,
        
          sizeof
        
        
          (G)); 
        
        
          for
        
         (
        
          int
        
         i=
        
          1
        
        ;i<=M;i++
        
          ) { 
        
        
          int
        
        
           u,v; scanf(
        
        
          "
        
        
          %d%d
        
        
          "
        
        ,&u,&
        
          v); G[u][v]
        
        =
        
          1
        
        
          ; G[v][u]
        
        =
        
          1
        
        
          ; } memset(color,
        
        -
        
          1
        
        ,
        
          sizeof
        
        
          (color)); flag
        
        =
        
          true
        
        
          ; 
        
        
          for
        
         (
        
          int
        
         i=
        
          1
        
        ;i<=N;i++
        
          ) 
        
        
          if
        
         (flag && color[i]==-
        
          1
        
        ) draw(i,
        
          0
        
        
          ); 
        
        
          if
        
         (!flag) printf(
        
          "
        
        
          No\n
        
        
          "
        
        
          ); 
        
        
          else
        
         printf(
        
          "
        
        
          %d\n
        
        
          "
        
        ,Max_match()/
        
          2
        
        
          ); } 
        
        
          return
        
        
          0
        
        
          ; }
        
      
    

?

The Accomodation of Students


更多文章、技術交流、商務合作、聯系博主

微信掃碼或搜索:z360901061

微信掃一掃加我為好友

QQ號聯系: 360901061

您的支持是博主寫作最大的動力,如果您喜歡我的文章,感覺我的文章對您有幫助,請用微信掃描下面二維碼支持博主2元、5元、10元、20元等您想捐的金額吧,狠狠點擊下面給點支持吧,站長非常感激您!手機微信長按不能支付解決辦法:請將微信支付二維碼保存到相冊,切換到微信,然后點擊微信右上角掃一掃功能,選擇支付二維碼完成支付。

【本文對您有幫助就好】

您的支持是博主寫作最大的動力,如果您喜歡我的文章,感覺我的文章對您有幫助,請用微信掃描上面二維碼支持博主2元、5元、10元、自定義金額等您想捐的金額吧,站長會非常 感謝您的哦!!!

發表我的評論
最新評論 總共0條評論
主站蜘蛛池模板: 久久高清免费 | 久久精品久久久久久久久人 | 在线观看高清国产福利视频 | 国产第一色 | 一区二区三区鲁丝不卡麻豆 | 99精品欧美一区二区三区 | 国产精品国产色综合色 | 亚洲产在线精品第一站不卡 | 久久精品免费i 国产 | 亚洲欧美中文字幕在线网站 | 欧美一级日韩 | 成人精品一级毛片 | 国产成人a大片大片在线播放 | 色久网站| 91日韩精品天海翼在线观看 | 麻豆首页| 性生生活三级视频观看 | 神马影院我不卡影院 | 亚洲视频一区二区三区四区 | 国产大片91精品免费看3 | www四虎影院 | 日本精品视频一区二区三区 | 视频一区国产 | 一级毛片免费高清视频 | 99久久99这里只有免费的精品 | 狠狠亚洲丁香综合久久 | 黄色小视频免费看 | 欧美精品免费在线观看 | 亚洲香蕉网综合久久 | 欧美一级毛片免费播放aa | 国产免费一区二区三区在线观看 | 欧美韩国日本一区 | 久久久久久综合 | 亚洲最大的成人网 | 亚洲久久草 | 92自拍视频 | 很黄很色的免费视频 | 欧美日本一级在线播放 | 国内精品伊人久久久影院 | 色老头久久久久久久久久 | 国产大片91精品免费观看不卡 |