ommand Network
| Time Limit: 1000MS | Memory Limit: 131072K | |
| Total Submissions: 20041 | Accepted: 5749 |
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
以下转自该博客：https://blog.csdn.net/shuangde800/article/details/8039359
题目大意：给定一个有向图，根节点已知，求该有向图的最小树形图。最小树形图即有向图的最小生成树，定义为：选择一些边，使得根节点能够到达图中所有的节点，并使得选出的边的边权和最小。
题目算法：朱-刘算法（即由中国人朱永津和刘振宏共同发明的算法）。
算法步骤如下：（本文不再证明，参考下面给出的我自己画的一个图即可理解）
1.判断图的连通性，若不连通直接无解，否则一定有解。
2.为除了根节点以外的所有点选择一个权值最小的入边，假设用pre数组记录前驱，f数组记录选择的边长，记所选边权和为temp。
3.（可利用并查集）判断选择的的边是否构成环，若没有则直接ans+=temp并输出ans，若有，则进行下一步操作。
4.对该环实施缩点操作，设该环上有点V1，V2……Vi……Vn，缩成的点为node ，对于所有不在环中的点P进行如下更改：
（1） 点P到node的距离为min{a[p,Vi]-f[Vi]} (a为边集数组)
（2）点node到p的距离为min{a[Vi,p]}
操作（1）的理解：先假设环上所有边均选上，若下次选择某一条边进入该环，则可以断开进入点与进入点的前驱之间的边，即断开F[进入点]，所以等效为直接把a[p,node]赋值为min{a[p,Vi]-f[Vi]}。
特别提醒：本题有自环，可以提前删掉，因为它没有用。
ac代码
/*
http://blog.csdn.net/liuke19950717
*/
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
const int inf=0x3f3f3f3f;
const int maxn=1005;
struct point
{
double x,y;
}p[maxn];
struct node
{
int u,v;
double len;
}edge[maxn*maxn];
int pre[maxn],id[maxn],vis[maxn];
double in[maxn];
double dis(point a,point b)
{
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
double dir_mst(int root,int n,int m)
{
double ans=0;
while(1)
{
//先找出所有点的最小入边
/*memset(in,inf,sizeof(in)); //这样写会报错！*/
for(int i=0;i<n;++i)
{
in[i]=inf;
}
for(int i=0;i<m;++i)
{
int u=edge[i].u,v=edge[i].v;
if(edge[i].len<in[v]&&u!=v)
{
pre[v]=u;in[v]=edge[i].len;
}
}
for(int i=0;i<n;++i)
{
if(i==root)
{
continue;
}
if(in[i]==inf)
{
return -1; //如果某点入度为零，必定找不到
}
}
//检查这些边是否构成了环
memset(id,-1,sizeof(id));
memset(vis,-1,sizeof(vis));
in[root]=0;
int cnt=0;
for(int i=0;i<n;++i)//标记环
{
ans+=in[i];
int v=i;
while(vis[v]!=i&&id[v]==-1&&v!=root)
{
vis[v]=i;
v=pre[v];
}
if(v!=root&&id[v]==-1)//缩点
{
for(int u=pre[v];u!=v;u=pre[u])
{
id[u]=cnt;
}
id[v]=cnt++;
}
}
if(cnt==0)
{
break;//无环
}
for(int i=0;i<n;++i)
{
if(id[i]==-1)
{
id[i]=cnt++;
}
}
//建立新图
for(int i=0;i<m;++i)
{
int u=edge[i].u,v=edge[i].v;
edge[i].u=id[u];
edge[i].v=id[v];
if(id[u]!=id[v])
{
edge[i].len-=in[v];
}
}
n=cnt;
root=id[root];
}
return ans;
}
int main()
{
int n,m;
while(~scanf("%d%d",&n,&m))
{
for(int i=0;i<n;++i)
{
scanf("%lf%lf",&p[i].x,&p[i].y);
}
for(int i=0;i<m;++i)
{
scanf("%d%d",&edge[i].u,&edge[i].v);
--edge[i].u;--edge[i].v;
if(edge[i].u!=edge[i].v)
{
edge[i].len=dis(p[edge[i].u],p[edge[i].v]);
}
else
{
edge[i].len=inf;
}
}
double ans=dir_mst(0,n,m);
if(ans==-1)
{
printf("poor snoopy\n");
}
else
{
printf("%.2f\n",ans);
}
}
return 0;
}
