# POJ 1639 Picnic Planning minimum spanning tree

2010-05-21  来源：本站原创  分类：Tech  人气：218

/ *
This title tune faster vomiting blood, and done really hard

This problem is mainly seeking minimum spanning tree, that is, given a tree, a node in the node with the maximum degree, then seek to meet such a condition the right of the tree and the final value and the minimum. This question is to do root Park limited degree, and then seek minimum spanning tree method for finding minimum spanning tree and the MST (minimum spanning tree closely related to) the main ideas in the following steps
1) to remove the root node from the graph
2) seek to remove the root node of the graph MST, pay attention to where the plan after the removal of the root may not be connected, so when the MST calculation should be calculated for each connected graph is where I chose to Kruscal algorithm + and inspection requirements set MST . After demand on each node connected in accordance with their respective component for coloring, and statistics the number of connected components
3) For each connected component, choose from the root node to the connected components in the node with the smallest weight edge to join the figure, assume that conn a connected component of the total need to add conn edges
4) from 1) 2) 3) steps that are a minimum spanning tree conn
5) Assume that subject qualified parking median kdeg, then the need to find kdeg - conn section starting from the root side, so cycle kdeg - conn times do the following:
a) Suppose the node k, edges [root, k, w] is not present in the tree (the right value of w), then by adding edges, if this tree will form a ring
b) assume that the ring is not directly connected with the root with the largest weight edge for the [from, to, weight]
Traverse all the tree nodes to find a maximum weight - w value of the edge, if the difference is greater than 0, will be this side [root, k] by adding trees and removal of side [from, to], if the difference value less than or equal 0 exit

This question is more frustrating feeling, ah, the code is mainly written in too many and too error prone to error two places
1) Calculate the MST when the calculation is wrong, the connected components of the color of different calculation confuse
2) the beginning edges of the array subscript set into 25, the result Runtime Error, then suddenly realized, 20 the maximum number of edges in the node is c (20, 2) far more than 25
It seems they still have to carefully point

6099743 bobten2008 1639 Accepted 252K 16MS C + + 5170B 2009-11-07 19:12:34
Over!
* /

```#include <iostream>
#include <algorithm>
#include <string>
#define MAX_N 25
using namespace std;
// Record all node name
struct node
{
string name;
int con[MAX_N + 1];     // Records the original point of the current point and the other side of the right values between
}nodes[MAX_N + 1];
int nodesn;                      // The number of nodes
int kdeg;                          // Minimal number of
struct edge                      // Used to record all the edges
{
int from, to, w;
edge()
{
from = to = w = 0;
}
}edges[MAX_N * MAX_N + 1];
int edgen;
int graph[MAX_N + 1][MAX_N + 1];
int color[MAX_N + 1];
int bfsq[MAX_N + 1], head, tail;
bool v[MAX_N + 1];
void init()
{
memset(nodes, 0, sizeof(nodes));
memset(graph, 0, sizeof(graph));
nodesn = 0;
edgen = 0;
}
int getIndex(const string &str)
{
for(int i = 0; i < nodesn; i++)
if(nodes[i].name == str)
return i;
return -1;
}
// Insert the side  [str1, str2, w]
void processNodes(const string &str1, const string &str2, int w)
{
int index1 = getIndex(str1);
if(index1 == -1)
{
index1 = nodesn++;
nodes[index1].name = str1;
}
int index2 = getIndex(str2);
if(index2 == -1)
{
index2 = nodesn++;
nodes[index2].name = str2;
}
nodes[index1].con[index2] = w;
nodes[index2].con[index1] = w;
if(str1 != "Park" && str2 != "Park")
{
edges[edgen].from = index1;
edges[edgen].to = index2;
edges[edgen].w = w;
edgen++;
}
}
bool compare(const edge &e1, const edge &e2)
{
return e1.w <= e2.w;
}
// Find sets of related
int sets[MAX_N + 1];
int find(int id)
{
if(sets[id] == id) return id;
else return sets[id] = find(sets[id]);
}
void joint(int id1, int id2)
{
int sid1 = find(id1), sid2 = find(id2);
if(sid1 == sid2) return;
else sets[sid1] = sid2;
}
//Kruskal  Algorithm,//return the number of Unicom component
int solveMST(int root)
{
int e = 0, t, colorseq = 0;
for(e = 0; e < nodesn; e++)
{
color[e] = 0;
sets[e] = e;
}
for(e = 0; e < edgen; e++)
{
int from = edges[e].from;
int to = edges[e].to;
int w = edges[e].w;
if(find(from) == find(to)) continue;
else
{
joint(from, to);
graph[from][to] = graph[to][from] = w;
}
}
// Coloring
for(e = 0; e < nodesn; e++)
{
if(color[e] || e == root) continue;
int sid = find(e);
color[e] = ++colorseq;
for(t = e + 1; t < nodesn; t++)
{
if(t == root) continue;
if(!color[t] && find(t) == sid)
color[t] = color[e];
}
}
return colorseq;
}
// Statistical records from the root to the    All points in the current tree path removed and  root Direct-attached to the edge of the right to the top edge of the   from to weight
void getMaxEdgeInPath(int root, int maxarray[])
{
head = tail = 1;
memset(bfsq, 0, sizeof(bfsq));
memset(v, 0, sizeof(v));
bfsq[tail] = root;
bfsq[tail] = 0;
tail = tail % MAX_N + 1;
v[root] = true;
while(head != tail)
{
int curId = bfsq[head];
int from = bfsq[head];
int to = bfsq[head];
int curMax = bfsq[head];
head = head % MAX_N + 1;
if(curId != root && from != root)
{
maxarray[curId] = from;
maxarray[curId] = to;
maxarray[curId] = curMax;
}
for(int toid = 0; toid < nodesn; toid++)
{
int curW;
if(v[toid] || (curW = graph[curId][toid]) == 0) continue;
v[toid] = true;
int fromm = from, too = to, curMaxx = curMax;
if(curW > curMax && curId != root)
{
fromm = curId;
too = toid;
curMaxx = curW;
}
bfsq[tail] = toid;
bfsq[tail] = fromm;
bfsq[tail] = too;
bfsq[tail] = curMaxx;
tail = tail % MAX_N + 1;
}
}
}
void solveKDegreeContraintTree()
{
// Records from the root to the    All points in the current tree path removed and  root Direct-attached to the edge of the right to the top edge of the   from to weight
int maxEdgeInPath[MAX_N + 1];
// Record point and root direct-attached
bool conDirectToRoot[MAX_N + 1];
// Record root and each connected component of weight between the smallest and weights
int minVal[MAX_N + 1];
memset(maxEdgeInPath, 0, sizeof(maxEdgeInPath));
memset(conDirectToRoot, 0, sizeof(conDirectToRoot));
memset(minVal, 0, sizeof(minVal));
int root = getIndex("Park");
int conn = solveMST(root);
int i, c;
// Select root and each connected component of weight between the smallest and weights
for(i = 0; i < nodesn; i++)
{
int w;
if(i == root || (w = nodes[root].con[i]) == 0) continue;
int c = color[i];
if(minVal[c] == 0 || w < minVal[c])
{
minVal[c] = i;
minVal[c] = w;
}
}
// Root and link  conn A connected component of the rooms has a minimum weight of the edge
for(c = 1; c <= conn; c++)
{
int to = minVal[c];
int w = minVal[c];
conDirectToRoot[to] = true;                   // Mark direct-attached
graph[root][to] = graph[to][root] = w;  // Even the edge
}
// Add extra kdeg-conn edges
int time = kdeg - conn;
while(time--)
{
memset(maxEdgeInPath, 0, sizeof(maxEdgeInPath));
getMaxEdgeInPath(root, maxEdgeInPath);
int minVal, minNode, maxsubval = 0;
// It loops through all of the nodes, look for the most weighted points difference  minNode
for(int k = 0; k < nodesn; k++)
{
int w;
if((w = nodes[root].con[k]) == 0 || conDirectToRoot[k]) continue;
if((maxEdgeInPath[k] - w) > maxsubval)
{
maxsubval = maxEdgeInPath[k] - w;
minVal = w;
minNode = k;
}
}
// No longer able to optimize, then exit
if(maxsubval == 0)
break;
// Replace the edge
int f = maxEdgeInPath[minNode], t = maxEdgeInPath[minNode];
graph[f][t] = graph[t][f] = 0;
graph[root][minNode] = graph[minNode][root] = minVal;
}

}
int main()
{
int in, i, j, w;
string from, to;
init();
cin>>in;
for(i = 0; i < in; i++)
{
cin>>from>>to>>w;
processNodes(from, to, w);
}
cin>>kdeg;
sort(edges, edges + edgen, compare);
solveKDegreeContraintTree();
int res = 0;
for(i = 0; i < nodesn; i++)
for(j = i + 1; j < nodesn; j++)
res += graph[i][j];
printf("Total miles driven: %d\n", res);
return 0;
}
```

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