1 #include <string.h>
2 #include<cstdio>
3 #include<stdio.h>
4 #include <iostream>
5 #include<string>
6 using namespace std;
7 int ll[1001][1001];
8 struct node
9 {
10 int per;
11 int sumlen;
12 bool vis;
13 node() :per(0), sumlen(0),vis(0) {};
14 };
15 node dd[1001];
16 int dijikstra(int s,int n)
17 {
18 int i,sum = 0,j, min,count=0;
19 while (count < n-1)
20 {
21 min = 1 << 30;
22 dd[s].vis = 1;//s为每次可能替换成的前驱
23 for (i = 1; i <= n; i++)
24 {
25 if (ll[s][i]>0&& !dd[i].vis)//次节点的前驱可修改
26 {
27 if (!dd[i].per)//没有前驱
28 {
29 dd[i].per = s;
30 dd[i].sumlen += ll[s][i]+dd[s].sumlen;
31 }
32 else//已有前驱 判断是否变动
33 {
34 if (dd[i].sumlen > dd[s].sumlen + ll[i][s])
35 {
36 dd[i].per = s;
37 dd[i].sumlen = dd[s].sumlen + ll[i][s];
38 }
39 else if (dd[i].sumlen == dd[s].sumlen + ll[i][s])//注意我们要求的是最短公路
40 {
41 if (ll[i][dd[i].per] > ll[i][s])
42 {
43 dd[i].per = s;
44 dd[i].sumlen = dd[s].sumlen + ll[i][s];
45 }
46 }
47 }
48 }
49 if (!dd[i].vis&&dd[i].sumlen&&min >=dd[i].sumlen)//min记录每次最短公路
50 {
51 if (min == dd[i].sumlen)
52 {
53 if (ll[dd[j].per][j] > ll[dd[i].per][i])
54 {
55 min = dd[i].sumlen;
56 j = i;
57 }
58 }
59 else
60 {
61 min = dd[i].sumlen;
62 j = i;
63 }
64 }
65 }
66 s = j;
67 sum += ll[dd[j].per][j];
68 count++;
69 }
70 return sum;
71 }
72 int main()
73 {
74 memset(ll,-1,sizeof(ll));
75 int n, m; cin >> n >> m;
76 //n = 4; m = 5;
77 int i, a, b, c;
78 for (i = 1; i <= n; i++)
79 ll[i][i] = 0;
80 for (i = 0; i < m; i++)
81 {
82 cin >> a >> b >> c;
83 ll[a][b] = ll[b][a] = c;
84 }
85 /*ll[1][2] = ll[2][1] = 4;
86 ll[1][3] = ll[3][1] = 5;
87 ll[2][3] = ll[3][2] = 2;
88 ll[2][4] = ll[4][2] = 3;
89 ll[3][4] = ll[4][3] = 2;*/
90 cout << dijikstra(1, n) << endl;
91 return 0;
92 }
