加权图是一种为每条边关联一个权值(可表示成本、时间等)的图模型。这种图能表示许多场景，如航空图中边表示航线，权值表示距离或费用。在航空图中，通常的问题是如何使距离或费用最小化。
我们可以通过加权无向图的最小生成树来解决这个问题。
图的生成树：是它的一颗含有其他所有顶点的无环连通子图。一幅加权无向图的最小生成树(MST)是它的一颗权值最小的生成树(树中所有边的权值之和最小)。
我们会一起学习计算最小生成树的两种经典算法：Prime算法和Kruskal算法。
首先有几个注意点：
- 只考虑连通图
- 边的权值可以表示距离、时间、费用或其他变量
- 边的权重可能是0或负数
- 所有边的权重都不相同
我们要在一幅加权连通无向图中找到它的最小生成树。
首先定义一下 加权无向图的数据结构
最小生成树算法：Prim算法和Kruskal算法
Prim算法：
Prim算法 模板：
1 #include <stdio.h> 2 #include <string.h> 3 #include <iostream> 4 #include <string> 5 #include <math.h> 6 #include <algorithm> 7 #include <vector> 8 #include <stack> 9 #include <queue> 10 #include <set> 11 #include <map> 12 #include <sstream> 13 const int INF=0x3f3f3f3f; 14 typedef long long LL; 15 const int mod=1e9+7; 16 //const double PI=acos(-1); 17 #define Bug cout<<"---------------------"<<endl 18 const int maxn=1e4+10; 19 using namespace std; 20 21 struct edge_node 22 { 23 int to; 24 int val; 25 int next; 26 }Edge[maxn*maxn/2]; 27 int Head[maxn]; 28 int tot; 29 30 void Add_Edge(int u,int v,int w) 31 { 32 Edge[tot].to=v; 33 Edge[tot].val=w; 34 Edge[tot].next=Head[u]; 35 Head[u]=tot++; 36 } 37 38 int lowval[maxn]; 39 int pre[maxn];//记录每个点的双亲是谁 40 41 int Prim(int n,int st)//n为顶点的个数,st为最小生成树的开始顶点 42 { 43 int sum=0; 44 memset(lowval,INF,sizeof(lowval)); 45 memset(pre,0,sizeof(pre)); 46 lowval[st]=-1; 47 pre[st]=-1; 48 for(int i=Head[st];i!=-1;i=Edge[i].next) 49 { 50 int v=Edge[i].to; 51 int w=Edge[i].val; 52 lowval[v]=w; 53 pre[v]=st; 54 } 55 for(int i=0;i<n-1;i++) 56 { 57 int MIN=INF; 58 int k; 59 for(int i=0;i<n;i++) 60 { 61 if(lowval[i]!=-1&&lowval[i]<MIN) 62 { 63 MIN=lowval[i]; 64 k=i; 65 } 66 } 67 sum+=MIN; 68 lowval[k]=-1; 69 for(int j=Head[k];j!=-1;j=Edge[j].next) 70 { 71 int v=Edge[j].to; 72 int w=Edge[j].val; 73 if(w<lowval[v]) 74 { 75 lowval[v]=w; 76 pre[v]=k; 77 } 78 } 79 } 80 return sum; 81 } 82 83 int main() 84 { 85 int n,m; 86 scanf("%d %d",&n,&m); 87 memset(Head,-1,sizeof(Head)); 88 for(int i=0;i<m;i++) 89 { 90 int u,v,w; 91 scanf("%d %d %d",&u,&v,&w); 92 Add_Edge(u,v,w); 93 Add_Edge(v,u,w); 94 } 95 printf("%d\n",Prim(n,0)); 96 return 0; 97 }
入门题：
Jungle Roads
http://poj.org/problem?id=1251
Description
The Head Elder of the tropical island of Lagrishan has a problem. A burst of foreign aid money was spent on extra roads between villages some years ago. But the jungle overtakes roads relentlessly, so the large road network is too expensive to maintain. The Council of Elders must choose to stop maintaining some roads. The map above on the left shows all the roads in use now and the cost in aacms per month to maintain them. Of course there needs to be some way to get between all the villages on maintained roads, even if the route is not as short as before. The Chief Elder would like to tell the Council of Elders what would be the smallest amount they could spend in aacms per month to maintain roads that would connect all the villages. The villages are labeled A through I in the maps above. The map on the right shows the roads that could be maintained most cheaply, for 216 aacms per month. Your task is to write a program that will solve such problems.
Input
The input consists of one to 100 data sets, followed by a final line containing only 0. Each data set starts with a line containing only a number n, which is the number of villages, 1 < n < 27, and the villages are labeled with the first n letters of the alphabet, capitalized. Each data set is completed with n-1 lines that start with village labels in alphabetical order. There is no line for the last village. Each line for a village starts with the village label followed by a number, k, of roads from this village to villages with labels later in the alphabet. If k is greater than 0, the line continues with data for each of the k roads. The data for each road is the village label for the other end of the road followed by the monthly maintenance cost in aacms for the road. Maintenance costs will be positive integers less than 100. All data fields in the row are separated by single blanks. The road network will always allow travel between all the villages. The network will never have more than 75 roads. No village will have more than 15 roads going to other villages (before or after in the alphabet). In the sample input below, the first data set goes with the map above.
Output
The output is one integer per line for each data set: the minimum cost in aacms per month to maintain a road system that connect all the villages. Caution: A brute force solution that examines every possible set of roads will not finish within the one minute time limit.
Sample Input
9 A 2 B 12 I 25 B 3 C 10 H 40 I 8 C 2 D 18 G 55 D 1 E 44 E 2 F 60 G 38 F 0 G 1 H 35 H 1 I 35 3 A 2 B 10 C 40 B 1 C 20 0
Sample Output
216 30
标准最小生成树，用了简化版的Prim算法，也可以用Kruskal算法
1 #include <stdio.h> 2 #include <string.h> 3 #include <algorithm> 4 #define maxn 65535 5 using namespace std; 6 int MG[1005][1005]; 7 int F[1005]; 8 9 int main() 10 { 11 //freopen("sample.txt","r",stdin); 12 int n; 13 while(~scanf("%d",&n)&&n) 14 { 15 int sum=0,i; 16 memset(MG,0,sizeof(MG)); 17 for(i=0;i<n-1;i++) 18 { 19 char c; 20 scanf(" %c",&c); 21 int a; 22 scanf("%d",&a); 23 for(int j=0;j<a;j++) 24 { 25 char k; 26 int b; 27 scanf(" %c",&k); 28 scanf("%d",&b); 29 MG[c-'A'][k-'A']=b; 30 MG[k-'A'][c-'A']=b; 31 } 32 F[i]=maxn; 33 } 34 F[i]=maxn; 35 F[0]=0; 36 for(i=1;i<n;i++) 37 { 38 if(MG[0][i]) 39 { 40 F[i]=MG[0][i]; 41 } 42 } 43 for(i=1;i<n;i++) 44 { 45 int m=maxn,t; 46 for(int g=0;g<n;g++) 47 { 48 if(F[g]&&F[g]<m) 49 { 50 t=g; 51 m=F[g]; 52 } 53 } 54 sum+=F[t]; 55 F[t]=0; 56 for(int j=1;j<n;j++) 57 { 58 if(F[j]&&MG[t][j]) 59 { 60 F[j]=min(MG[t][j],F[j]); 61 } 62 } 63 } 64 printf("%d\n",sum); 65 } 66 return 0; 67 }