Networking （最小生成树）

You are assigned to design network connections between certain points in a wide area. You are given a set of points in the area, and a set of possible routes for the cables that may connect pairs of points. For each possible route between two points, you are given the length of the cable that is needed to connect the points over that route. Note that there may exist many possible routes between two given points. It is assumed that the given possible routes connect (directly or indirectly) each two points in the area. Your task is to design the network for the area, so that there is a connection (direct or indirect) between every two points (i.e., all the points are interconnected, but not necessarily by a direct cable), and that the total length of the used cable is minimal.

Input

The input file consists of a number of data sets. Each data set defines one required network. The first line of the set contains two integers: the first defines the number P of the given points, and the second the number R of given routes between the points. The following R lines define the given routes between the points, each giving three integer numbers: the first two numbers identify the points, and the third gives the length of the route. The numbers are separated with white spaces. A data set giving only one number P=0 denotes the end of the input. The data sets are separated with an empty line. The maximal number of points is 50. The maximal length of a given route is 100. The number of possible routes is unlimited. The nodes are identified with integers between 1 and P (inclusive). The routes between two points i and j may be given as i j or as j i.

Output

For each data set, print one number on a separate line that gives the total length of the cable used for the entire designed network.

Sample Input

1 0 2 3 1 2 37 2 1 17 1 2 68 3 7 1 2 19 2 3 11 3 1 7 1 3 5 2 3 89 3 1 91 1 2 32 5 7 1 2 5 2 3 7 2 4 8 4 5 11 3 5 10 1 5 6 4 2 12 0

Sample Output

0 17 16 26 #include <cstdio> #include <iostream> #include <cmath> #include <string> #include <cstring> #include <algorithm> #include <queue> using namespace std; #define ll long long const int maxn = 100+8; const int inf = 0x3f3f3f; int p, r, sign[maxn]; struct point//用结构体，更容易对边进行遍历 { int u, v, w; }root[58]; int init()//使每个点在初始不与任何点相连 { for(int i = 0; i<maxn; i++) sign[i] = i; } int get(int x)//寻找祖先 { if(sign[x] == x)//判断x点是否为末节点 return x; else//如果不是末节点，那就找它的祖先 { sign[x] = get(sign[x]); return sign[x]; } } int bcj(int x, int y)//看x点与y点是否同一个祖先 { int t1 = get(x); int t2 = get(y); if(t1 == t2) return 0;//如果x点与y点同一个祖先，那就不需要任何操作 else//如果不同祖先，则使他们变成同一个祖先 { sign[t1] = t2; return 1; } } int cmp(point a, point b)//使边从小到大排序 { return a.w<b.w; } int main() { while(~scanf("%d", &p) && p) { scanf("%d", &r); init(); int sum = 0, cnt = 0; for(int i = 0; i<r; i++) { scanf("%d%d%d", &root[i].u, &root[i].v, &root[i].w); } sort(root, root+r, cmp);//使边从小到大排序 for(int i = 0; i<r; i++) { if(bcj(root[i].u, root[i].v))//判断u与v是否同一个祖先，判断成功则为不同祖先 { cnt++; sum += root[i].w;//既然不同祖先，就让他们的边相加 } if(cnt == p-1)break;//最小生成树只有n-1条边 } printf("%d\n", sum); } return 0; }

 

转载于:https://www.cnblogs.com/RootVount/p/10513333.html