Networking POJ - 1287 （简单最小生成树）
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
题意 ：给定两个点及其边权值，求出各点权值和最小的值（即生成最小生成树）
解题方法：最小生成树
这道题是一棵裸的最小二叉树，即克鲁斯卡尔算法（kruskal），算法大概内容是，首先对所有的边权值进行排序，并判断两个顶点是否连通，判断方法可借鉴并查集思想，将所有的点放入并查集中，判断两个顶点是否连通，即判断顶点是否属于一个集合内，以提升算法效率。可以用邻接矩阵存储边权值，不过在相同点之间存在多个权值时，需要取最小的权。
代码：
// Kruskal
#include
#include
#include
using namespace std;
struct edge{
int u;
int v;
int w;
}ed[5001];
int n,m;
int f[5001] = {0};
int getf(int v){
if(f[v] == v)
return v;
else{
f[v] = getf(f[v]);
return f[v];
}
}
int merge(int u,int v){
int t1,t2;
t1 = getf(u);
t2 = getf(v);
if(t1 != t2){
f[t2] = t1;
return 1;
}
return 0;
}
bool cmp(edge ed1,edge ed2){
return ed1.w < ed2.w;
}
int main(){
int n,m;
while(scanf("%d",&n) && n != 0){
cin >> m;
int sum = 0,count = 0;
for(int i = 1; i <= m; i++){
scanf("%d%d%d",&ed[i].u,&ed[i].v,&ed[i].w);
}
sort(ed + 1, ed + m + 1, cmp);
for(int i = 1; i <= n; i++)
f[i] = i;
for(int i = 1; i <= m; i++){
if(merge(ed[i].u, ed[i].v)){
count++;
sum += ed[i].w;
}
if(count == n - 1)
break;
}
printf("%d\n",sum);
}
return 0;
}