HDU1102--Constructing Roads(最小生成树)

Posted 2020-08-25 Pic

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

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 }