• <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 3164 Command Network 最小樹形圖

            Description

            After a long lasting war on words, a war on arms finally breaks out between littleken’s and KnuthOcean’s kingdoms. A sudden and violent assault by KnuthOcean’s force has rendered a total failure of littleken’s command network. A provisional network must be built immediately. littleken orders snoopy to take charge of the project.

            With the situation studied to every detail, snoopy believes that the most urgent point is to enable littenken’s commands to reach every disconnected node in the destroyed network and decides on a plan to build a unidirectional communication network. The nodes are distributed on a plane. If littleken’s commands are to be able to be delivered directly from a node A to another node B, a wire will have to be built along the straight line segment connecting the two nodes. Since it’s in wartime, not between all pairs of nodes can wires be built. snoopy wants the plan to require the shortest total length of wires so that the construction can be done very soon.

            Input

            The input contains several test cases. Each test case starts with a line containing two integer N (N ≤ 100), the number of nodes in the destroyed network, and M (M ≤ 104), the number of pairs of nodes between which a wire can be built. The next N lines each contain an ordered pair xi and yi, giving the Cartesian coordinates of the nodes. Then follow M lines each containing two integers i and j between 1 and N (inclusive) meaning a wire can be built between node i and node j for unidirectional command delivery from the former to the latter. littleken’s headquarter is always located at node 1. Process to end of file.

            Output

            For each test case, output exactly one line containing the shortest total length of wires to two digits past the decimal point. In the cases that such a network does not exist, just output ‘poor snoopy’.

            Sample Input

            4 6
            0 6
            4 6
            0 0
            7 20
            1 2
            1 3
            2 3
            3 4
            3 1
            3 2
            4 3
            0 0
            1 0
            0 1
            1 2
            1 3
            4 1
            2 3

            Sample Output

            31.19
            poor snoopy

            Source


             

            最小樹形圖算法(Zhu-Liu Algorithm)

            1.       設最小樹形圖的總權值為cost,置cost0

            2.       除源點外,為其他所有節點Vi找一條權值最小的入邊,加入集合TT就是最短邊的集合。加邊的方法:遍歷所有點到Vi的邊中權值最小的加入集合T,記pre[Vi]為該邊的起點,mincost[Vi]為該邊的權值。

            3.       檢查集合T中的邊是否存在有向環,有則轉到步驟4,無則轉到步驟5。這里需要利用pre數組,枚舉檢查過的點作為搜索的起點,類似dfs的操作判斷有向環。

            4.       將有向環縮成一個點。設環中有點{Vk1,Vk2,…,Vki}i個點,用Vk代替縮成的點。在壓縮后的圖中,更新所有不在環中的點VVk的距離:

            map[V][Vk] = min {map[V][Vkj]-mincost[Vki]} 1<=j<=i

            map[Vk][V] = min {map[Vkj][V]}           1<=j<=I

            5.       cost加上T中有向邊的權值總和就是最小樹形圖的權值總和。

            #include <iostream>
            #include 
            <cmath>

            #define min(a,b) (a<b ? a:b)

            const int MAXN = 110;
            const int INF = 0x7FFFFFFF;
            int n,m,pre[MAXN];
            double x[MAXN],y[MAXN];
            bool circle[MAXN],visit[MAXN];
            double ans,map[MAXN][MAXN];

            inline 
            double distance(double x1,double y1,double x2,double y2){
                
            return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
            }

            void dfs(int u){
                visit[u]
            =true;
                
            for(int i=2;i<=n;i++)
                    
            if(!visit[i] && map[u][i]!=INF)
                        dfs(i);
            }

            bool connected(){
                memset(visit,
            false,sizeof(visit));
                
            int i,cnt=0;
                
            for(i=1;i<=n;i++)
                    
            if(!visit[i])
                        dfs(i),cnt
            ++;
                
            return cnt==1 ? true : false;
            }

            void min_arborescence(){
                
            int i,j,k;
                memset(circle,
            false,sizeof(circle));
                
            while(true){
                    
            for(i=2;i<=n;i++){
                        
            if(circle[i]) continue;
                        pre[i]
            =i;
                        map[i][i]
            =INF;
                        
            for(j=1;j<=n;j++){
                            
            if(circle[j]) continue;
                            
            if(map[j][i]<map[pre[i]][i])
                                pre[i]
            =j;
                        }

                    }

                    
            for(i=2;i<=n;i++){
                        
            if(circle[i]) continue;
                        j
            =i;
                        memset(visit,
            false,sizeof(visit));
                        
            while(!visit[j] && j!=1){
                            visit[j]
            =true;
                            j
            =pre[j];
                        }

                        
            if(j==1continue;
                        i
            =j;
                        ans
            +=map[pre[i]][i];
                        
            for(j=pre[i];j!=i;j=pre[j]){
                            ans
            +=map[pre[j]][j];
                            circle[j]
            =true;
                        }

                        
            for(j=1;j<=n;j++){
                            
            if(circle[j]) continue;
                            
            if(map[j][i]!=INF)
                                map[j][i]
            -=map[pre[i]][i];
                        }

                        
            for(j=pre[i];j!=i;j=pre[j])
                            
            for(k=1;k<=n;k++){
                                
            if(circle[k]) continue;
                                map[i][k]
            =min(map[i][k],map[j][k]);
                                
            if(map[k][j]!=INF)
                                    map[k][i]
            =min(map[k][i],map[k][j]-map[pre[j]][j]);
                            }

                        
            break;
                    }

                    
            if(i>n){
                        
            for(j=2;j<=n;j++){
                            
            if(circle[j]) continue;
                            ans
            +=map[pre[j]][j];
                        }

                        
            break;
                    }

                }

            }

            int main(){
                
            int i,j,u,v;
                
            while(scanf("%d %d",&n,&m)!=EOF){
                    
            for(ans=i=0;i<=n;i++for(j=0;j<=n;j++) map[i][j]=INF;
                    
            for(i=1;i<=n;i++) scanf("%lf %lf",&x[i],&y[i]);
                    
            while(m--){
                        scanf(
            "%d %d",&u,&v);
                        map[u][v]
            =distance(x[u],y[u],x[v],y[v]);
                    }

                    
            if(!connected()) puts("poor snoopy");
                    
            else{
                        min_arborescence();
                        printf(
            "%.2lf\n",ans);
                    }

                }

                
            return 0;
            }

            posted on 2009-05-26 16:03 極限定律 閱讀(677) 評論(0)  編輯 收藏 引用 所屬分類: ACM/ICPC

            <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

            亚洲中文字幕无码久久2017| 香蕉99久久国产综合精品宅男自| 亚洲欧美伊人久久综合一区二区 | 91精品国产91久久久久福利| 久久久婷婷五月亚洲97号色| 久久久久久狠狠丁香| 国产精品美女久久久久av爽| 无码国内精品久久综合88| 久久精品国产亚洲77777| 久久免费视频一区| 久久久久人妻精品一区三寸蜜桃| 国产高潮国产高潮久久久91| 久久久久se色偷偷亚洲精品av| 久久亚洲综合色一区二区三区| 一极黄色视频久久网站| 久久精品国产亚洲7777| 麻豆AV一区二区三区久久| 国产真实乱对白精彩久久| 亚洲精品国产字幕久久不卡| 热久久国产欧美一区二区精品| 无码人妻久久久一区二区三区| 中文字幕一区二区三区久久网站| 亚洲精品无码久久久久去q| 婷婷久久综合| 久久av高潮av无码av喷吹| 色综合合久久天天综合绕视看| 久久婷婷国产剧情内射白浆 | 欧美伊人久久大香线蕉综合| 国色天香久久久久久久小说| 久久亚洲AV无码西西人体| 精品久久久久久综合日本| 久久综合88熟人妻| 人人妻久久人人澡人人爽人人精品 | 伊人久久大香线蕉AV一区二区| 久久电影网2021| 国产午夜精品久久久久免费视| 久久久噜噜噜久久中文字幕色伊伊| 国产精品VIDEOSSEX久久发布| 9久久9久久精品| 蜜桃麻豆www久久| 久久96国产精品久久久|