HDU 1301 最小生成树，以及最小生成树的学习

这个题我用的是Kruskal算法

1.概览

Kruskal算法是一种用来寻找最小生成树的算法，由Joseph Kruskal在1956年发表。用来解决同样问题的还有Prim算法和Boruvka算法等。三种算法都是贪婪算法的应用。和Boruvka算法不同的地方是，Kruskal算法在图中存在相同权值的边时也有效。

 

2.算法简单描述

1).记Graph中有v个顶点，e个边

2).新建图Graph_new，Graph_new中拥有原图中相同的e个顶点，但没有边

3).将原图Graph中所有e个边按权值从小到大排序

4).循环：从权值最小的边开始遍历每条边 直至图Graph中所有的节点都在同一个连通分量中

                if 这条边连接的两个节点于图Graph_new中不在同一个连通分量中

                                         添加这条边到图Graph_new中

还有一种算法，还没有学会，学会后来补上这个！！

The Head Elder of the tropical island of Lagrishan has a problem. A burst of foreign aid money was spent on extra roads between villages some years ago. But the jungle overtakes roads relentlessly, so the large road network is too expensive to maintain. The Council of Elders must choose to stop maintaining some roads. The map above on the left shows all the roads in use now and the cost in aacms per month to maintain them. Of course there needs to be some way to get between all the villages on maintained roads, even if the route is not as short as before. The Chief Elder would like to tell the Council of Elders what would be the smallest amount they could spend in aacms per month to maintain roads that would connect all the villages. The villages are labeled A through I in the maps above. The map on the right shows the roads that could be maintained most cheaply, for 216 aacms per month. Your task is to write a program that will solve such problems. 

The input consists of one to 100 data sets, followed by a final line containing only 0. Each data set starts with a line containing only a number n, which is the number of villages, 1 < n < 27, and the villages are labeled with the first n letters of the alphabet, capitalized. Each data set is completed with n-1 lines that start with village labels in alphabetical order. There is no line for the last village. Each line for a village starts with the village label followed by a number, k, of roads from this village to villages with labels later in the alphabet. If k is greater than 0, the line continues with data for each of the k roads. The data for each road is the village label for the other end of the road followed by the monthly maintenance cost in aacms for the road. Maintenance costs will be positive integers less than 100. All data fields in the row are separated by single blanks. The road network will always allow travel between all the villages. The network will never have more than 75 roads. No village will have more than 15 roads going to other villages (before or after in the alphabet). In the sample input below, the first data set goes with the map above. 

The output is one integer per line for each data set: the minimum cost in aacms per month to maintain a road system that connect all the villages. Caution: A brute force solution that examines every possible set of roads will not finish within the one minute time limit. 

Input

9A 2 B 12 I 25B 3 C 10 H 40 I 8C 2 D 18 G 55D 1 E 44E 2 F 60 G 38F 0G 1 H 35H 1 I 353A 2 B 10 C 40B 1 C 200

Output

21630

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <algorithm>#include <vector>#include <map>#include <math.h>#include <stack>#define LL long longusing namespace std;const int INF = 0x3f3f3f3f;const int maxn = 10000 + 10;int dir[4][2] = {{1,0},{0,1},{-1,0},{0,-1}};int cur;struct edges{    int fr,en,val;}edge[maxn];int par[maxn];bool cmp(edges a,edges b){    return a.val < b.val;}void init(){    for(int i = 0;i < maxn;i++)    {        par[i] = i;    }}int find(int x){    if(x == par[x])return x;    return par[x] = find(par[x]);}int kruskal(){    int sum = 0;    for(int i = 0;i < cur;i++)    {        int nx = find(edge[i].fr);        int ny = find(edge[i].en);        if(nx != ny)        {            par[nx] = ny;            //用来输出个边及其权值            //cout<<char(edge[i].fr+'A')<<" "<<char(edge[i].en + 'A' )<<" "<<edge[i].val<<endl;            sum+=edge[i].val;//求和        }    }    return sum;}int main(){    int n;    while(cin>>n,n)    {        init();        cur = 0;        for(int i= 0;i < n-1;i++)        {            char st,hou;            int t,p;            cin>>st>>t;            for(int i = 0;i < t;i++)            {                cin>>hou>>p;                edge[cur].fr = st - 'A';                edge[cur].en = hou - 'A';                edge[cur++].val = p;            }        }        sort(edge,edge+cur,cmp);        int k = kruskal();        cout<<k<<endl;    }    return 0;}