PE 107 Minimal network （MST）

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最新推荐文章于 2024-11-09 14:42:29 发布

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分类专栏： Project Euler ACM_最小生成树文章标签： PE 107 Minimal network MST

本文链接：https://blog.youkuaiyun.com/liangzhaoyang1/article/details/53846824

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本文探讨了一个经典的图论问题——最小生成树(MST)，通过具体实例介绍如何从一个加权无向图中找到总权重最小的子图，确保所有顶点仍保持连通，并给出了一种实现Prim算法的C++代码，用于解决实际问题。

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Minimal network

Problem 107

The following undirected network consists of seven vertices and twelve edges with a total weight of 243.

The same network can be represented by the matrix below.

	A	B	C	D	E	F	G
A	-	16	12	21	-	-	-
B	16	-	-	17	20	-	-
C	12	-	-	28	-	31	-
D	21	17	28	-	18	19	23
E	-	20	-	18	-	-	11
F	-	-	31	19	-	-	27
G	-	-	-	23	11	27	-

However, it is possible to optimise the network by removing some edges and still ensure that all points on the network remain connected. The network which achieves the maximum saving is shown below. It has a weight of 93, representing a saving of 243 − 93 = 150 from the original network.

Using network.txt (right click and 'Save Link/Target As...'), a 6K text file containing a network with forty vertices, and given in matrix form, find the maximum saving which can be achieved by removing redundant edges whilst ensuring that the network remains connected.

题解：Minimum Spanning Tree...

代码：

#include<bits/stdc++.h>
#define MIN INT_MAX
#define MAX_Point 1200   //最大顶点数 
#define MAX_Edge 144000  //最大的边数 
using namespace std;
typedef long long ll;
int flag1 =0;
ll sum;
ll arr_list[MAX_Point][MAX_Point]; 
struct Edge
{
	ll point;
	ll lowcost;
	ll flag;
};
Edge edge[MAX_Edge];
ll prim(int n)
{
	ll i,j,k=1,flag;
	ll min,sum2=0;
	j=1;
	for(i=1;i<=n;i++)
	{
		if(i!=j)
		{
			edge[i].point=i;
			edge[i].lowcost=arr_list[j][i];
			edge[i].flag=0;
		}
	}
	edge[j].lowcost=0;
	edge[j].flag=1;
	for(i=2;i<=n;i++)
	{
		k=1;
		min=MIN;
		flag=0;
		for(j=2;j<=n;j++)
		{
			if(edge[j].flag==0 && edge[j].lowcost<min)
			{
				k=j;
				min=edge[j].lowcost;
				flag=1;
			}
		}
		if(!flag) return -1;
		
		sum2+=min;
		edge[k].flag=1;
		for(j=2;j<=n;j++)
		{
			if(edge[j].flag==0 && arr_list[k][j]<edge[j].lowcost)
			{
				edge[j].point=k;
				edge[j].lowcost=arr_list[k][j];
			}
		}
	}
	return sum2;
}

int main()
{
	
	ll sum0=0;
	freopen("in.txt","r",stdin);
	for(int i=1;i<=40;i++)
	{
		for(int j=1;j<=40;j++)
		{
			cin>>arr_list[i][j];
			sum0 += arr_list[i][j];
			if(arr_list[i][j]==0)
			arr_list[i][j]=0xfffffff;
		//	if(arr_list=='-')
		}
	}
	sum0/=2;
	ll ans=0;
	ans=prim(40);
	cout<<sum0<<endl;
	cout<<sum0-ans<<endl;
	return 0;
}
/*
0 16 12 21 0 0 0
16 0 0 17 20 0 0
12 0 0 28 0 31 0
21 17 28 0 18 19 23
0 20 0 18 0 0 11
0 0 31 19 0 0 27
0 0 0 23 11 27 0
*/