Problem Description
There are N
villages, which are numbered from 1 to N, and you should build some
roads such that every two villages can connect to each other. We
say two village A and B are connected, if and only if there is a
road between A and B, or there exists a village C such that there
is a road between A and C, and C and B are connected.
We know that there are already some roads between some villages and
your job is the build some roads such that all the villages are
connect and the length of all the roads built is
minimum.
Input
The first line
is an integer N (3 <= N <= 100), which is the number of
villages. Then come N lines, the i-th of which contains N integers,
and the j-th of these N integers is the distance (the distance
should be an integer within [1, 1000]) between village i and
village j.
Then there is an integer Q (0 <= Q <= N * (N + 1) / 2). Then
come Q lines, each line contains two integers a and b (1 <= a
< b <= N), which means the road between village a and village
b has been built.
Output
You should
output a line contains an integer, which is the length of all the
roads to be built such that all the villages are connected, and
this value is minimum.
Sample Input
3 0 990 692
990 0 179 692 179 0 1 1 2
Sample Output
题意：给你一个N*N的矩阵内容是，从I到j的距离，让你求出使这个城镇联通需要建造的最短公路数；
解题思路：这个题和D题都是kruskal，刚开始用kruskal的时候自己想出来的，后来才知道这是一种算法0.0，天才，嘻嘻嘻~
感悟：并查集的模板有点多，得多看看啊
代码：
#include
#define M 120
#define INF 0x3f3f3f3f
using namespace std;
int bin[M],m[M][M];
struct node
{
int i,j,w;//i,j表示的是从i到j，需要w的距离
};
node fr[M*M];
bool comp(node &a,node &b)
{
return a.w
}
int findx(int x)//找根
{
int r=x;
while(bin[r]!=r)
r=bin[r];
return r;
}
int kruskal(int n,int m)
{
sort(fr,fr+m,comp);//将需要的距离从小到大排列
int ans=0;
for(int i=0;i
{
int fx=findx(fr[i].i);
int fy=findx(fr[i].j);
//cout<<fx<<" "<<fy<<endl;
if(fx!=fy)
{
ans+=fr[i].w;
bin[fx]=fy;
//cout<<"YES"<<endl;
}
}
return ans;
}
int main()
{
//freopen("in.txt","r",stdin);
int n,t;
while(scanf("%d",&n)!=EOF)
{
for(int i=1;i<=n;i++)
bin[i]=i;
int m=0,k;
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
{
scanf("%d",&k);
//cout<<k<<endl;
if(i>=j) continue;
fr[m].i=i;
fr[m].j=j;
fr[m++].w=k;
}
scanf("%d",&t);
int a,b;
for(int i=1;i<=t;i++)
{
scanf("%d%d",&a,&b);
int fx=findx(a);
int fy=findx(b);
bin[fx]=fy;
}
//for(int i=1;i<=n;i++)
//
cout<<bin[i]<<" ";
//cout<<endl;
//cout<<m<<endl;
int ans=kruskal(n,m);
printf("%d\n",ans);
}
return 0;
}
