Command Network
| Time Limit: 1000MS | Memory Limit: 131072K | |
| Total Submissions: 15914 | Accepted: 4583 |
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 Mlines 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
题意：
给出一个有向图，给出n个顶点的坐标，以及每条边连接的顶点，指定一个起点，找到一个方案，使得从这个点到其他所有点的路径的权值和最小，求最小权值
题解：
传说中的最小树形图，相当于无向图的最小生成树，但是考虑的东西多了很多，毕竟有向图的连通性判断都比较复杂了....
不是太会这个算法，大概看明白了点，正在理解中，大神博客点这里...
网上找不到太多相关的内容，想试试模板都难....
/*
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;
}