HDU 1102  Constructing Roads

Constructing Roads

Time Limit: 2000/1000 MS
(Java/Others)    Memory Limit:
65536/32768 K (Java/Others)
Total Submission(s): 4902
   Accepted Submission(s):
1757

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
179
 

Source
kicc
 

Recommend
Eddy
典型的最小生成树，这种题原来做过多少遍了，可是还不知道这就叫最小生成树


先把有效的路径从小到大排序，然后合并村庄，不在一个father节点的说明要合并，加上他的距离，然后合并，最后的距离就是所求的结果啦，嘿嘿，没想到1A
代码：

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
struct point{
int begin,end,len;
}dist[10000];
int mark[108];
int cmp(const void *a,const void *b)
{
struct point *c,*d;
c=(struct point *)a;
d=(struct point *)b;
return c->len-d->len;
}
int father(int x)
{
if(mark[x]==x)
return x;
mark[x]=father(mark[x]);
return mark[x];
}
int main()
{
int N,i,j,k,Q,a,b,fa,fb,sum;
int flag[108][108];
while(scanf("%d",&N)!=EOF)
{
memset(flag,0,sizeof(flag));
k=0;
for(i=1;i<=N;i++)
for(j=1;j<=N;j++)
{
scanf("%d",&a);
if(i<j){
dist[k].len=a;
dist[k].begin=i;
dist[k++].end=j;
}
}
for(i=0;i<=N;i++)
mark[i]=i;
scanf("%d",&Q);
for(i=0;i<Q;i++)
{
scanf("%d%d",&a,&b);
fa=father(a);
fb=father(b);
if(fa!=fb){
if(fa>fb)
mark[fa]=fb;
else mark[fb]=fa;
}
}
qsort(dist,k,sizeof(dist[0]),cmp);
sum=0;
for(i=0;i<k;i++)
{
fa=father(dist[i].begin);
fb=father(dist[i].end);
if(fa!=fb){
sum+=dist[i].len;
if(fa>fb)
mark[fa]=fb;
else mark[fb]=fa;
}
}
printf("%d\n",sum);
}
return 0;
}