• <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

            精品久久国产一区二区三区香蕉| 亚洲国产精品成人AV无码久久综合影院| 久久久久久亚洲精品成人| 久久精品国产99久久久| 国产精品成人99久久久久91gav| 天天综合久久一二三区| 久久一日本道色综合久久| 国产精品无码久久久久| 热re99久久6国产精品免费| 久久99精品久久久久久秒播| 欧美黑人激情性久久| 狠狠色伊人久久精品综合网| 久久天天躁狠狠躁夜夜躁2O2O| 久久久久亚洲精品中文字幕| 欧美丰满熟妇BBB久久久| 亚洲国产成人精品91久久久 | 午夜视频久久久久一区| 精品一区二区久久久久久久网站| 国产69精品久久久久APP下载| 久久91精品综合国产首页| 久久天堂电影网| 国产精品一久久香蕉国产线看观看 | 精品国产乱码久久久久软件| 天天久久狠狠色综合| 97精品国产91久久久久久| 亚洲AV无码久久精品成人| 久久亚洲AV成人无码| 精品综合久久久久久97| 热99RE久久精品这里都是精品免费| 狠狠色综合网站久久久久久久| 久久青草国产精品一区| 99久久伊人精品综合观看| 日韩一区二区久久久久久| 久久久久免费精品国产| 日韩一区二区久久久久久 | 无码任你躁久久久久久| 久久精品国产亚洲AV不卡| 欧美精品福利视频一区二区三区久久久精品 | 亚洲愉拍99热成人精品热久久| 国产A级毛片久久久精品毛片| 久久强奷乱码老熟女网站|