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;
}