• <ins id="pjuwb"></ins>
    <blockquote id="pjuwb"><pre id="pjuwb"></pre></blockquote>
    <noscript id="pjuwb"></noscript>
          <sup id="pjuwb"><pre id="pjuwb"></pre></sup>
            <dd id="pjuwb"></dd>
            <abbr id="pjuwb"></abbr>

            POJ 3177 Redundant Paths 雙連通分量+縮點

            Description

            In order to get from one of the F (1 <= F <= 5,000) grazing fields (which are numbered 1..F) to another field, Bessie and the rest of the herd are forced to cross near the Tree of Rotten Apples. The cows are now tired of often being forced to take a particular path and want to build some new paths so that they will always have a choice of at least two separate routes between any pair of fields. They currently have at least one route between each pair of fields and want to have at least two. Of course, they can only travel on Official Paths when they move from one field to another.

            Given a description of the current set of R (F-1 <= R <= 10,000) paths that each connect exactly two different fields, determine the minimum number of new paths (each of which connects exactly two fields) that must be built so that there are at least two separate routes between any pair of fields. Routes are considered separate if they use none of the same paths, even if they visit the same intermediate field along the way.

            There might already be more than one paths between the same pair of fields, and you may also build a new path that connects the same fields as some other path.

            Input

            Line 1: Two space-separated integers: F and R

            Lines 2..R+1: Each line contains two space-separated integers which are the fields at the endpoints of some path.

            Output

            Line 1: A single integer that is the number of new paths that must be built.

            Sample Input

            7 7
            1 2
            2 3
            3 4
            2 5
            4 5
            5 6
            5 7

            Sample Output

            2

            Hint

            Explanation of the sample:

            One visualization of the paths is:
               1   2   3
            +---+---+
            | |
            | |
            6 +---+---+ 4
            / 5
            /
            /
            7 +
            Building new paths from 1 to 6 and from 4 to 7 satisfies the conditions.
               1   2   3
            +---+---+
            : | |
            : | |
            6 +---+---+ 4
            / 5 :
            / :
            / :
            7 + - - - -
            Check some of the routes:
            1 – 2: 1 –> 2 and 1 –> 6 –> 5 –> 2
            1 – 4: 1 –> 2 –> 3 –> 4 and 1 –> 6 –> 5 –> 4
            3 – 7: 3 –> 4 –> 7 and 3 –> 2 –> 5 –> 7

            Every pair of fields is, in fact, connected by two routes.

            It's possible that adding some other path will also solve the problem (like one from 6 to 7). Adding two paths, however, is the minimum.

            Source


                題意大意:一群牛將被在一個特定路徑構成的農場上遷移,每兩塊農場之間都至少有一條通道,這些牛要求每兩塊路徑至少要有兩條通道,求最少需要修建多少條路才能滿足要求。
                這題的解法與http://www.shnenglu.com/mythit/archive/2009/05/29/86082.html完全一樣,只是題目中說了圖中有可能存在平行邊,這里必須判斷一下。我還是很偷懶的用了STL里的vector模擬鄰接矩陣,并且開了個5001*5001的bool數組判斷平行邊。結果導致代碼的效率和空間消耗都很大,110MS和將近24M的內存空間。如果自己建圖的話,效率能提高很多。

            #include <iostream>
            #include 
            <vector>
            using namespace std;

            const int MAXN = 5001;
            vector
            < vector<int> > adj;
            bool hash[MAXN][MAXN];
            int cnt,low[MAXN],pre[MAXN],visit[MAXN],degree[MAXN];

            void dfs(int u,int v){
                visit[u]
            =1;
                pre[u]
            =cnt++,low[u]=pre[u];
                
            int i,len=adj[u].size();
                
            for(i=0;i<len;i++){
                    
            if(adj[u][i]==v) continue;
                    
            if(!visit[adj[u][i]]) dfs(adj[u][i],u);
                    
            if(low[adj[u][i]]<low[u]) low[u]=low[adj[u][i]];
                }

                visit[u]
            =2;
            }

            int main(){
                
            int i,j,u,v,n,m,len,ans;
                
            while(scanf("%d %d",&n,&m)!=EOF){
                    adj.assign(n
            +1,vector<int>());
                    memset(hash,
            false,sizeof(hash));
                    
            while(m--){
                        scanf(
            "%d %d",&u,&v);
                        
            if(!hash[u][v]){
                            hash[u][v]
            =true;
                            adj[u].push_back(v),adj[v].push_back(u);
                        }

                    }

                    memset(visit,
            0,sizeof(visit));
                    cnt
            =0,dfs(1,1);
                    memset(degree,
            0,sizeof(degree));
                    
            for(i=1;i<=n;i++){
                        len
            =adj[i].size();
                        
            for(j=0;j<len;j++)
                            
            if(low[i]!=low[adj[i][j]])
                                degree[low[i]]
            ++;
                    }

                    
            for(ans=i=0;i<=n;i++)
                        
            if(degree[i]==1) ans++;
                    printf(
            "%d\n",(ans+1)/2);
                }

                
            return 0;
            }

            posted on 2009-05-30 01:18 極限定律 閱讀(1546) 評論(4)  編輯 收藏 引用 所屬分類: ACM/ICPC

            評論

            # re: POJ 3177 Redundant Paths 雙連通分量+縮點 2009-08-14 09:53 zeus

            省去hash可以這樣判重空間小很多 時間沒多多少 依然0ms
            bool isok( int u, int v )//判重
            {
            for ( int i= 0; i< g[u].size(); ++i )
            if ( g[u][i]== v ) return false;

            return true;
            }  回復  更多評論   

            # re: POJ 3177 Redundant Paths 雙連通分量+縮點 2009-08-14 20:55 極限定律

            我也想這樣做的,不過怕時間效率變低,就偷懶直接HASH了@zeus
              回復  更多評論   

            # re: POJ 3177 Redundant Paths 雙連通分量+縮點 2011-04-28 09:30 Icyeye

            拜讀了哈,幫助很大,謝啦^-^
            但是有一點,那個visit[u]=2不知道有什么用,但注釋掉后能快三分之二左右的時間~~  回復  更多評論   

            # re: POJ 3177 Redundant Paths 雙連通分量+縮點[未登錄] 2012-07-31 20:48 bigrabbit

            樓主,我發現個問題。這組數據對于下面的數據
            5 6
            1 2
            1 3
            2 3
            3 4
            3 5
            4 5
            輸出的low數組是 0 0 0 1 1
            是不對的,應該是0 0 0 0 0,你建圖的方式很奇怪,我也看不懂你到底是怎么建圖的??梢越忉屜聠??我直接用vector<int> edg[]搞的,刪除重邊。  回復  更多評論   

            <2009年5月>
            262728293012
            3456789
            10111213141516
            17181920212223
            24252627282930
            31123456

            導航

            統計

            常用鏈接

            留言簿(10)

            隨筆分類

            隨筆檔案

            友情鏈接

            搜索

            最新評論

            閱讀排行榜

            評論排行榜

            色狠狠久久综合网| 中文无码久久精品| 久久电影网2021| 无码久久精品国产亚洲Av影片 | 久久国产高清字幕中文| 久久精品www| 久久影院亚洲一区| 久久婷婷国产剧情内射白浆| 久久久久人妻一区精品色 | 日本精品久久久久中文字幕| 久久强奷乱码老熟女网站| 亚洲AV无码久久| 欧美日韩中文字幕久久久不卡| 久久久久久伊人高潮影院| 久久亚洲精品视频| 欧美精品国产综合久久| 亚洲国产成人久久精品动漫| 亚洲国产小视频精品久久久三级| 久久国产热精品波多野结衣AV| 久久精品一区二区影院| 国内精品久久久久伊人av| 日日狠狠久久偷偷色综合96蜜桃 | 久久精品九九亚洲精品| 久久国产免费| 国产一区二区三区久久精品| 久久午夜福利无码1000合集| 精品久久人人爽天天玩人人妻| 久久久噜噜噜久久熟女AA片| 久久久久久午夜精品| 久久综合精品国产一区二区三区 | 婷婷久久精品国产| 精品久久久久久国产三级| 国产午夜福利精品久久2021 | 亚洲精品乱码久久久久久按摩| 亚洲国产综合久久天堂| 精品久久久无码中文字幕| 日韩精品久久久久久| 久久综合九色综合精品| 国产一久久香蕉国产线看观看| 国产精品久久网| 久久本道伊人久久|