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.
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Input The first line is an integer N (3 <= N <= 100), which is Then |
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Output You should output a line contains an integer, which is the length |
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Sample Input 3 0 990 692 990 0 179 692 179 0 1 1 2 |
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Sample Output 179 |
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Source kicc |
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Recommend Eddy |
最小生成树的模板题目
下面的代码使用了Prim算法
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<algorithm>
5 #include<cmath>
6 #include<map>
7 #include<iomanip>
8 #include<queue>
9 #define INF 0x7ffffff
10 #define MAXN 200
11 using namespace std;
12 const double eps=1e-10;
13 const double PI=acos(-1);
14 int G[MAXN][MAXN];
15 int vnew[MAXN];
16 int lowval[MAXN];
17 int sum;
18 int n,q;
19 void Prim(int start)
20 {
21 int j,mi;
22 for(int i=1;i<=n;i++){
23 if(i!=start){
24 lowval[i]=G[start][i];
25 vnew[i]=0;
26 }
27 }
28 vnew[start]=1;
29 for(int i=1;i<=n-1;i++){
30 j=-1;
31 mi=INF;
32 for(int i=1;i<=n;i++){
33 if(vnew[i]==0&&lowval[i]<mi){
34 j=i;
35 mi=lowval[i];
36 }
37 }
38 vnew[j]=1;
39 sum+=lowval[j];
40 for(int i=1;i<=n;i++){
41 if(vnew[i]==0){
42 lowval[i]=min(lowval[i],G[j][i]);
43 }
44 }
45 }
46 }
47 int main()
48 {
49 #ifndef ONLINE_JUDGE
50 freopen("data.in", "r", stdin);
51 #endif
52 std::ios::sync_with_stdio(false);
53 std::cin.tie(0);
54 //Prim算法
55 int a,b;
56 while(cin>>n){
57 for(int i=1;i<=n;i++){
58 for(int j=1;j<=n;j++){
59 cin>>G[i][j];
60 }
61 }
62 cin>>q;
63 for(int i=0;i<q;i++){
64 cin>>a>>b;
65 G[a][b]=G[b][a]=0;
66 }
67 sum=0;
68 Prim(1);
69 cout<<sum<<endl;
70 }
71
72 }