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hdu 1102 Constructing Roads （最小生成树）
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1102
Constructing Roads
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14172 Accepted Submission(s): 5402
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.
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.
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
题目大意：第一行表示n*n的矩阵，然后输入矩阵，表示的就是1到1的距离，1到2的距离。。。。第二行是2到1的距离，2到2的距离。。。。以此类推~
然后在输入一行，表示下面有几组数据，下面输入的两个数就是表示两个之间有路，不需要再建了，最后就是输出能够连通所有道路的最短路了！！！
一个最小生成树prim就可以解决了！哇哈哈~
1 #include <iostream> 2 #include <cstdio> 3 using namespace std; 4 int map[1010][1010],node[1010],Min,n,a,b; 5 const int INF=999999999; 6 7 int prim() 8 { 9 int vis[1010]= {0};10 int tm=1,m,s=0;11 vis[tm]=1;12 node[tm]=0;13 for (int k=1; k<=n; k++)14 {15 Min=INF;16 for (int i=1; i<=n; i++)17 if (!vis[i])18 {19 if (node[i]>map[tm][i])20 node[i]=map[tm][i];21 if (Min>node[i])22 {23 Min=node[i];24 m=i;25 }26 //s+=Min;27 }28 tm=m;29 vis[m]=1;30 31 }32 for (int i=1; i<=n; i++)33 s+=node[i];34 return s;35 }36 37 int main ()38 {39 while (cin>>n)40 {41 //cout<<n<<endl;42 for (int i=1; i<=n; i++)43 {44 node[i]=INF;45 for (int j=1; j<=n; j++)46 map[i][j]=INF;47 }48 for (int i=1; i<=n; i++)49 {50 for (int j=1; j<=n; j++)51 {52 cin>>map[i][j];53 }54 }55 int q;56 cin>>q;57 while (q--)58 {59 cin>>a>>b;60 map[a][b]=map[b][a]=0;61 }62 printf ("%d\n",prim());63 }64 return 0;65 }
hdu 1102 Constructing Roads （最小生成树）
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