标签：des style blog class code java
Constructing Roads
Time Limit: 2000/1000 MS
(Java/Others) Memory Limit: 65536/32768 K
(Java/Others)
Total Submission(s): 12967 Accepted
Submission(s): 4911
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
Source
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入门级题目：
1 //31MS 328K 1150B G++
2 #include<stdio.h>
3 #include<stdlib.h>
4 struct node{
5 int u;
6 int v;
7 int d;
8 }p[10005];
9 int set[105],n;
10 int cmp(const void*a,const void*b)
11 {
12 return (*(node*)a).d-(*(node*)b).d;
13 }
14 inline int find(int x)
15 {
16 if(x!=set[x]) set[x]=find(set[x]);
17 return set[x];
18 }
19 inline int merge(int x,int y)
20 {
21 int a=find(x);
22 int b=find(y);
23 if(a!=b){
24 set[a]=b;
25 return 1;
26 }
27 return 0;
28 }
29 inline int kruskal()
30 {
31 int ans=0;
32 for(int i=0;i<n*n;i++)
33 if(merge(p[i].u,p[i].v)){
34 ans+=p[i].d;
35 }
36 return ans;
37 }
38 int main(void)
39 {
40 int m,u,v;
41 while(scanf("%d",&n)!=EOF)
42 {
43 int k=0;
44 for(int i=0;i<=n;i++) set[i]=i;
45 for(int i=1;i<=n;i++)
46 for(int j=1;j<=n;j++){
47 scanf("%d",&p[k].d);
48 p[k].u=i;
49 p[k].v=j;
50 k++;
51 }
52 scanf("%d",&m);
53 while(m--){
54 scanf("%d%d",&u,&v);
55 merge(u,v);
56 }
57 qsort(p,n*n,sizeof(p[0]),cmp);
58 printf("%d\n",kruskal());
59 }
60 return 0;
61 }
hdu 1102 Constructing Roads (最小生成树),布布扣,bubuko.com
hdu 1102 Constructing Roads (最小生成树)
标签：des style blog class code java
原文地址：http://www.cnblogs.com/GO-NO-1/p/3709970.html
