POJ3164 Command Network【最小树形图】【朱刘算法】
Command Network
Time Limit: 1000MS Memory Limit: 131072K
Total Submissions: 13398Accepted: 3868
DescriptionAfter 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
31.19
poor snoopy
POJ Monthly--2006.12.31, galaxy
题目大意：长时间战争过后，一场战争终于切断了Littleken和KnuthOcean王国的联系。
Littleken的指挥网络瘫痪了，现在最重要的事是建立一个临时的通信网络，这个任务交
给了Snoopy。
Snoopy觉得最重要的一点是要把命令传到被摧毁的网络中的每一个点上，所以他决定先
建立一个单向的传输网络。假设所有的传输节点都分布在一个平面上。如果Littleken的
命令想要从节点A传送到节点B上，必须建立一个单向电缆从节点A连接到节点B。为了
尽可能节省资源，要求通信网络所用的电缆长度最小(参考ACM-ICPC程序设计系列)。
现在给你N个点坐标(x,y)，M条可以架设有向电缆的路。求：至少需要的电缆长度。如果
不能建立这样的通信网，就输出"poor snoopy"。
思路：理解题目意思后，就转换为给你一个有向图，求有向图的最小树形图。在这里要用
到朱刘算法(终于见到我们中国人自己写出的算法了)。具体步骤如下：
朱刘算法(Edmonds)：
基于贪心和缩点的思想。
假设根的顶点是V0。
(1)除了根结点外，所有的点Vi，找到以Vi为终点的最短的边，加入集合中
（pre[v]存放的是终点v的起点,In[v]存放终点为v的最短的边）
(2)检查集合中有没有有向环和收缩点。若没有有向环和收缩点，结束计算；若没有有向环、
但含收缩边，则跳至步骤(4)；若含有有向环，则跳至步骤(3)。
(3)含有有向环，则收缩有向环(缩点)，把有向环收缩为一个点，其有向环内的边被收缩掉，而环外的边被保
留，构建新图，重复步骤(1)、(2)。
(4)没有有向环，有收缩边，则展开收缩边。
来个图还是很有必要的~~~
#include
#include
#include
#include
#include
#include
using namespace std;
const int MAXN = 110;
const int MAXM = 10010;
struct Node
{
int from;
int to;
double w;
};
Node Edges[MAXM];
struct Node1
{
double x;
double y;
};
Node1 Point[MAXN];
double Dist(Node1 a, Node1 b)
{
double x = a.x - b.x;
double y = a.y - b.y;
return sqrt(x*x+y*y);
}
int pre[MAXN],vis[MAXN],flag[MAXN];
double In[MAXN],sum;
double ZhuLiu(int root,int N,int M)
{
sum = 0;
while(true)
{
for(int i = 0; i < N; ++i)
In[i] = INT_MAX;
for(int i = 0; i < M; ++i)
{
int u = Edges[i].from;
int v = Edges[i].to;
if(Edges[i].w < In[v] && u != v)
{
pre[v] = u;//v为终点，pre[v]存放起点
In[v] = Edges[i].w;//权值最小的边
}
}
for(int i = 0; i < N; ++i)//如果存在除root以外的孤立点，则不存在最小树形图
{
if(i == root)
continue;
if(In[i] == INT_MAX)
return -1;
}
int CntNode = 0;
memset(flag,-1,sizeof(flag));
memset(vis,-1,sizeof(vis));
In[root] = 0;
for(int i = 0; i < N; ++i) //找环,标记每个环
{
sum += In[i];
int v = i;
while(vis[v]!=i && flag[v]==-1 && v!=root)//每个点寻找其前序点，要么最终寻找至根部，要么找到一个环
{
vis[v] = i;
v = pre[v];
}
if(v != root && flag[v] == -1) //新图重新编号
{
for(int u = pre[v]; u != v; u = pre[u])
flag[u] = CntNode;
flag[v] = CntNode++;
}
}
if(CntNode == 0) //无环，跳出
break;
for(int i = 0; i < N; ++i)
{
if(flag[i] == -1)
flag[i] = CntNode++;
}
for(int i = 0; i < M; ++i) //建立新图，更新其他点到环的距离
{
int v = Edges[i].to;
Edges[i].from = flag[Edges[i].from];
Edges[i].to = flag[Edges[i].to];
if(Edges[i].from != Edges[i].to)
Edges[i].w -= In[v];
}
N = CntNode;
root = flag[root];
}
return sum;
}
int main()
{
int x,y,N,M;
while(~scanf("%d%d",&N,&M))
{
int id = 0;
for(int i = 0; i < N; ++i)
scanf("%lf%lf",&Point[i].x,&Point[i].y);
for(int i = 0; i < M; ++i)
{
scanf("%d%d",&x,&y);
if(x == y)
continue;
x--;
y--;
Edges[id].from = x;
Edges[id].to = y;
Edges[id++].w = Dist(Point[x],Point[y]);
}
double ans = ZhuLiu(0,N,id);
if(ans == -1)
printf("poor snoopy\n");
else
printf("%.2lf\n",ans);
}
return 0;
}