hdu1102 Constructing Roads (簡單最小產生樹Prim演算法)

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Problem DescriptionThere 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. 

 

InputThe 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. 

 

OutputYou 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 Input30 990 692990 0 179692 179 011 2 

 

Sample Output179  很簡單的一個最小產生樹 題意也很簡單 看完直接AC了  好開心O(∩_∩)O~~

 1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5  6 using namespace std; 7  8 #define inf 99999999 9 10 int map[105][105],path,node[105],vis[105],n;11 12 void Prim()13 {14     int ans=1;15     vis[1]=1;16     node[1]=0;17     for(int k=1;k<=n;k++)18     {19         int minn=inf;20         for(int i=1;i<=n;i++)21         if(!vis[i]&&minn>node[i])22         {23             minn=node[i];24             ans=i;25         }26         vis[ans]=1;27         path+=node[ans];28 29         for(int i=1;i<=n;i++)30         {31             if(!vis[i]&&node[i]>map[ans][i])32             node[i]=map[ans][i];33         }34     }35 }36 37 int main()38 {39     while(cin>>n)40     {41         for(int i=1;i<=n;i++)42         {43             node[i]=inf;vis[i]=0;44             for(int j=1;j<=n;j++)45             cin>>map[i][j];46         }47         int m;48         cin>>m;49         while(m--)50         {51             int a,b;52             cin>>a>>b;53             map[a][b]=map[b][a]=0;54         }55         path=0;56         Prim();57         cout<<path<<endl;58     }59     return 0;60 }

 

hdu1102 Constructing Roads (簡單最小產生樹Prim演算法)