POJ 3164 Command Network 最小树形图

POJ 3164 Command Network 最小树形图

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 x_i and y_i, 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

POJ Monthly--2006.12.31, galaxy

 

最小树形图算法(Zhu-Liu Algorithm)：

1.       设最小树形图的总权值为cost，置cost为0。

2.       除源点外，为其他所有节点Vi找一条权值最小的入边，加入集合T。T就是最短边的集合。加边的方法：遍历所有点到Vi的边中权值最小的加入集合T，记pre[Vi]为该边的起点，mincost[Vi]为该边的权值。

3.       检查集合T中的边是否存在有向环，有则转到步骤4，无则转到步骤5。这里需要利用pre数组，枚举检查过的点作为搜索的起点，类似dfs的操作判断有向环。

4.       将有向环缩成一个点。设环中有点{Vk1,Vk2,…,Vki}共i个点，用Vk代替缩成的点。在压缩后的图中，更新所有不在环中的点V到Vk的距离：

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中有向边的权值总和就是最小树形图的权值总和。

#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==1) continue;
            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;
}