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 }