题意：求连接所有村子的最短路
prim+邻接矩阵
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <cstdlib>
5 #include <algorithm>
6 using namespace std;
7 #define inf 999999
8 #define N 30
9 int n,dis[N],vis[N],g[N][N];
10
11 int prim(int root)
12 {
13 for(int i=1; i<=n; i++)
14 dis[i]= g[i][root],vis[i]= 0;
15 dis[root]= 0;
16 vis[root]=1;
17 int cost= 0;
18 for(int T=1; T<=n-1; T++)
19 {
20 int mindis= inf,idx= -1;
21 for(int i=1; i<=n; i++)
22 {
23 if( vis[i]==0 )
24 {
25 if( mindis>dis[i] )
26 {
27 mindis= dis[i];
28 idx= i;
29 }
30 }
31 }
32 cost+=mindis;
33 if( idx==-1 ) return -1;
34 vis[idx]= 1;
35 for(int i=1; i<=n; i++)
36 {
37 if( vis[i]==0 )
38 {
39 if( dis[i]>g[i][idx] )
40 dis[i]= g[i][idx];
41 }
42 }
43 }
44 return cost;
45 }
46
47 int main()
48 {
49 char op1[2],op2[2];
50 int co,w;
51 while(scanf("%d",&n)&&n)
52 {
53 for(int i=0; i<=N; i++)
54 for(int j=0; j<=N; j++)
55 {
56 if(i==j) g[i][j] = 0;
57 else g[i][j] = inf;
58 }
59 for(int i=1; i<=n-1; i++)
60 {
61 scanf("%s%d",op1,&co);
62 for(int j=1; j<=co; j++)
63 {
64 scanf("%s%d",op2,&w);
65 int u = op1[0]-'A'+1;
66 int v = op2[0]-'A'+1;
67 g[u][v] = g[v][u] = min(g[u][v],w);
68 }
69 }
70 int ans=prim(1);
71 printf("%d\n",ans);
72 }
73 return 0;
74 }
prim+优先队列
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<cstdlib>
5 #include<algorithm>
6 #include<queue>
7 using namespace std;
8 #define M 200
9 #define N 30
10 const int inf= 999999;
11 int n,size;
12 typedef pair<int,int>mp;
13
14 int head[N];
15 int dis[N];
16 int vis[N];
17
18 struct Edge
19 {
20 int v,w,next;
21 Edge(){}
22 Edge(int V,int W,int NEXT):v(V),w(W),next(NEXT){}
23 }edge[M];
24
25 void Init()
26 {
27 memset(head,-1,sizeof(head));
28 size = 0;
29 }
30
31 void InsertEdge(int u,int v,int w)
32 {
33 edge[size] = Edge(v,w,head[u]);
34 head[u] = size++;
35 edge[size] = Edge(u,w,head[v]);
36 head[v] =size++;
37 }
38
39 int prim(int s)
40 {
41 int cost= 0;
42 int times = 0;
43 for(int i=1; i<=n; i++)
44 {
45 vis[i] = 0;
46 dis[i] = inf;
47 }
48 mp p(0,s);
49 priority_queue< mp,vector<mp>,greater<mp> > Q;
50 Q.push(mp(0,s));
51 while(!Q.empty()&×<n)
52 {
53 do
54 {
55 p = Q.top();
56 Q.pop();
57 }while(vis[p.second]==1&&!Q.empty());
58 if(vis[p.second]==0)
59 {
60 cost += p.first;
61 vis[p.second] = 1;
62 times ++;
63 for(int i = head[p.second]; i!=-1; i=edge[i].next)
64 {
65 int v = edge[i].v;
66 if(vis[v]==0)
67 {
68 if(edge[i].w < dis[v])
69 {
70 dis[v] = edge[i].w;
71 Q.push(mp(dis[v],v));
72 }
73 }
74 }
75 }
76 }
77 if(times<n)
78 return -1;
79 return cost;
80 }
81 int main()
82 {
83 char op1[2],op2[2];
84 int co,w;
85 while(scanf("%d",&n)&&n)
86 {
87 Init();
88 for(int i=1; i<=n-1; i++)
89 {
90 scanf("%s%d",op1,&co);
91 for(int j=1; j<=co; j++)
92 {
93 scanf("%s%d",op2,&w);
94 int u = op1[0]-'A'+1;
95 int v = op2[0]-'A'+1;
96 int flag = 0;
97 for(int k=head[u]; k!=-1; k=edge[k].next)
98 {//重边的情况
99 if(v == edge[j].v)
100 {
101 if(w < edge[j].w)
102 edge[j].w = w;
103 flag = 1;
104 break;
105 }
106 }
107 if(flag) continue;
108 InsertEdge(u,v,w);
109 }
110 }
111 int ans=prim(1);
112 printf("%d\n",ans);
113 }
114 return 0;
115 }
kruskal
1 #include <iostream>
2 #include <cstdio>
3 #include <cstdlib>
4 #include <cstring>
5 #include <algorithm>
6 using namespace std;
7
8 #define N 30
9 #define M 100
10 int n,m,father[N],rank[N],sum;
11 struct Edge
12 {
13 int u;
14 int v;
15 int w;
16 }edge[M];
17
18 bool cmp(Edge a,Edge b)
19 {
20 return a.w < b.w ;
21 }
22
23 void Init()
24 {
25 for(int i=0; i<=n; i++)
26 {
27 father[i] = i;
28 rank[i] = 1;
29 }
30 }
31
32 int find(int x)
33 {
34 if(father[x] != x)
35 {
36 father[x] = find(father[x]);
37 }
38 return father[x];
39 }
40
41 void merge(int x,int y)
42 {
43 int xf = find(x);
44 int yf = find(y);
45 if(rank[x] > rank[y])
46 {
47 rank[xf] += rank[yf];
48 father[yf] = xf;
49 }
50 else
51 {
52 rank[yf] += rank[xf];
53 father[xf] = yf;
54 }
55 }
56
57 void Kruskal()
58 {
59 int num = 0;
60 for(int i=0; i<m; i++)
61 {
62 int u = edge[i].u;
63 int v = edge[i].v;
64 if(find(u) != find(v))
65 {
66 num++;
67 sum += edge[i].w;
68 merge(u,v);
69 }
70 if(num==n-1) break;
71 }
72 }
73
74 int main()
75 {
76 char op1[2],op2[2];
77 int co,w,cnt;
78 while(scanf("%d",&n)&&n)
79 {
80 Init();
81 cnt = 0;
82 for(int i=1; i<n; i++)
83 {
84 scanf("%s%d",op1,&co);
85 for(int j=1; j<=co; j++)
86 {
87 scanf("%s%d",op2,&w);
88 int u = op1[0]-'A'+1;
89 int v = op2[0]-'A'+1;
90 edge[cnt].u = u ; edge[cnt].v = v ; edge[cnt].w = w;
91 cnt++;
92 }
93 }
94 m = cnt;
95 sort(edge,edge+m,cmp);
96 sum = 0;
97 Kruskal();
98 printf("%d\n",sum);
99 }
100 return 0;
101 }
来源：https://www.cnblogs.com/ar940507/p/3223357.html