题意：给出一个无向图，求在已知顶点v0的度不超过K的情况下，所得的最小生成树
题解：
首先不考虑v0点，先求得v1-v(n-1)的MST，然后分两种情况考虑：
令d为v0的度
代码：
#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <string>
#include <map>
using namespace std;
const int V = 21;
const int INF = 1 << 30;
int n;
struct Edge
{
int from;
int to;
int weight;
};
map<string,int> Map;
int graph[V][V];
Edge edges[V];
int vertexNum;
int s;
bool visited[V];//边(0,i)是否在edges中
void init()
{
memset(visited,0,sizeof(visited));
Map.clear();
for(int i = 0; i < V; ++i)
{
for(int j = 0; j < V; ++j)
{
graph[i][j] = INF;
}
}
}
void input()
{
string name1,name2;
int dis;
Map["Park"] = 0;
int k = 1;
for(int i = 0; i < n; ++i)
{
cin >> name1 >> name2 >> dis;
if(Map.find(name1) == Map.end())
Map[name1] = k++;
if(Map.find(name2) == Map.end())
Map[name2] = k++;
int id1 = Map[name1];
int id2 = Map[name2];
graph[id1][id2] = graph[id2][id1] = dis;
}
scanf("%d",&s);
}
//求v0 - v(_vertexNum-1)的最小生成树
int Prim(int _vertexNum)
{
int mstWeight = 0;
for(int i = 1; i < _vertexNum - 1; ++i)
{
edges[i].from = 1;
edges[i].to = i + 1;
edges[i].weight = graph[1][i+1];
}
for (int i = 2; i < _vertexNum; ++i)
{
int id = i-1;
int minW = edges[i-1].weight;
for (int j = i; j < _vertexNum-1; ++j)
{
if (minW > edges[j].weight)
{
minW = edges[j].weight;
id = j;
}
}
mstWeight += minW;
swap(edges[i-1],edges[id]);
int k = edges[i-1].to;
for (int j = i; j < _vertexNum -1; ++j)
{
int v = edges[j].to;
int w = graph[k][v];
if(w < edges[j].weight)
{
edges[j].weight = w;
edges[j].from = k;
}
}
}
return mstWeight;
}
//返回回路中最大的边
bool isCycle;
void maxWeightInCycle(int _mv,int _from,int _to,int& _maxW,int& _id)
{
if (_to == _mv)
{
isCycle = true;
return;
}
for (int i = 0; i < vertexNum-1; ++i)
{
if (edges[i].from != _to && edges[i].to != _to)
{
continue;
}
if (edges[i].from == _to && edges[i].to != _from)
{
maxWeightInCycle(_mv,_to,edges[i].to,_maxW,_id);
if (isCycle)
{
if (_maxW < edges[i].weight && edges[i].to != 0)
{
_maxW = edges[i].weight;
_id = i;
}
break;
}
}
else if(edges[i].to == _to && edges[i].from != _from)
{
maxWeightInCycle(_mv,_to,edges[i].from,_maxW,_id);
if (isCycle)
{
if (_maxW < edges[i].weight && edges[i].from != 0)
{
_maxW = edges[i].weight;
_id = i;
}
break;
}
}
}
}
void Test()
{
init();
input();
vertexNum = Map.size();
int ans = Prim(vertexNum);//v0 - vn-1
int minW = INF;
int id = -1;
for (int i = 1; i < vertexNum; ++i)
{
if (minW > graph[0][i])
{
minW = graph[0][i];
id = i;
}
}
ans += graph[0][id];
visited[id] = true;
//添加边(0,id)
edges[0].from = 0;
edges[0].to = id;
edges[0].weight = minW;
//枚举顶点0 的度
for (int d = 2; d <= s; ++d)
{
int dec = INF;
int edgeId = -1;
id = -1;
for (int i = 1; i < vertexNum; ++i)
{
if (visited[i])//已经在MST树中了
{
continue;
}
int maxW = 0,maxId = -1;
isCycle = false;
//添加边(0-i)，并返回回路最大边
maxWeightInCycle(0,0,i,maxW,maxId);
if (dec > graph[0][i] - maxW)
{
dec = graph[0][i] - maxW;
edgeId = maxId;
id = i;
}
}
if (dec >= 0)
{
break;
}
else
{
//将回路最大边删除，并修改为0点的边
if (edgeId != -1)
{
edges[edgeId].from = 0;
edges[edgeId].to = id;
edges[edgeId].weight = graph[0][id];
}
visited[id] = true;
ans += dec;
}
}
printf("Total miles driven: %d\n",ans);
}
int main()
{
//freopen("data.txt","r",stdin);
while(scanf("%d",&n) != EOF)
{
Test();
}
return 0;
}posted on 2011-06-02 11:49
bennycen 阅读(1359)
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