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(shè)最小樹形圖的總權(quán)值為cost，置cost為0。
2. 除源點(diǎn)外，為其他所有節(jié)點(diǎn)Vi找一條權(quán)值最小的入邊，加入集合T。T就是最短邊的集合。加邊的方法：遍歷所有點(diǎn)到Vi的邊中權(quán)值最小的加入集合T，記pre[Vi]為該邊的起點(diǎn)，mincost[Vi]為該邊的權(quán)值。
3. 檢查集合T中的邊是否存在有向環(huán)，有則轉(zhuǎn)到步驟4，無則轉(zhuǎn)到步驟5。這里需要利用pre數(shù)組，枚舉檢查過的點(diǎn)作為搜索的起點(diǎn)，類似dfs的操作判斷有向環(huán)。
4. 將有向環(huán)縮成一個(gè)點(diǎn)。設(shè)環(huán)中有點(diǎn){Vk1,Vk2,…,Vki}共i個(gè)點(diǎn)，用Vk代替縮成的點(diǎn)。在壓縮后的圖中，更新所有不在環(huán)中的點(diǎn)V到Vk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(quán)值總和就是最小樹形圖的權(quán)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==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;
}