1 //本代码解决，一个图的最小生成树的各边权值总和
2
3 #include<iostream>
4 using namespace std;
5 //static声明静态变量，只运行一次， const限定一个变量不允许被改变，产生静态作用。
6 static const int MAX = 100;
7 static const int INFTY = (1<<21);
8 static const int WHITE = 0;
9 static const int GRAY = 1;
10 static const int BLACK = 2;
11
12 int n, M[MAX][MAX];
13
14 int prim() {
15 int u, minv;
16 /*
17 1.d[v]记录连接内顶点与V - T 内顶点的边中，权值最小的边的权值
18 2.p[v]记录MST中顶点的父节点
19 */
20 int d[MAX], p[MAX], color[MAX];
21
22 for(int i = 0; i < n; i++) {
23 d[i] = INFTY;
24 p[i] = -1;
25 color[i] = WHITE;
26 }
27
28 d[0] = 0;
29
30 while(1) {
31 minv = INFTY;
32 u = -1;
33 for(int i = 0; i < n; i++) {
34 if(minv > d[i] && color[i] != BLACK){
35 u = i;
36 minv = d[i];
37 }
38 }
39 //所有点都被遍历结束循环
40 if(u == -1) break;
41
42 color[u] = BLACK;
43 //更新与之相关的顶点
44 for(int v = 0; v < n; v++) {
45 if(color[v] != BLACK && M[u][v] != INFTY) {
46 if(d[v] > M[u][v]) {
47 d[v] = M[u][v];
48 p[v] = u;
49 color[v] = GRAY;
50 }
51 }
52 }
53 }
54
55 int sum = 0;
56 for(int i = 0; i < n; i++) {
57 if(p[i] != -1) sum += M[i][p[i]]; //这里为什么不用sum+=d[i]呢，因为害怕它不是连通图
58 }
59 return sum;
60 }
61
62 int main() {
63 cin >> n;
64 //构建邻接矩阵
65 for(int i = 0; i < n; i++) {
66 for(int j = 0; j < n; j++) {
67 int e;
68 cin >> e;
69 M[i][j] = (e == -1) ? INFTY: e;
70 }
71 }
72
73 cout << prim() << endl;
74 return 0;
75 }
76 /*
77 5
78 -1 2 3 1 -1
79 2 -1 -1 4 -1
80 3 -1 -1 1 1
81 1 4 1 -1 3
82 -1 -1 1 3 -1
83 */
