HDU1102--Constructing Roads-白红宇

HDU1102--Constructing Roads

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发布时间：2019-06-14

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Constructing Roads

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 10598 Accepted Submission(s): 3942

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

0 990 692

990 0 179

692 179 0

1 2

Sample Output

179

Source

kicc

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1 #include
      
         2 #include
       
          3 #include
        
           4 #include
         
            5 #include
          
             6 #include
            7 using namespace std; 8 9 int parent[5001]; 10 int n; 11 int edge[1001][1001]; 12 13 struct node 14 { 15 int u,v; 16 int w; 17 }point[5001]; 18 19 bool cmp(node a,node b) 20 { 21 return a.w
            
             0;s=parent[s]){} 37 while(s!=x) 38 { 39 int temp=parent[x]; 40 parent[x]=s; 41 x=temp; 42 } 43 return s; 44 } 45 46 void Union(int R1,int R2) 47 { 48 int r1=Find(R1),r2=Find(R2); 49 int temp=parent[r1]+parent[r2]; 50 if(parent[r1]>parent[r2]) 51 { 52 parent[r1]=r2; 53 parent[r2]=temp; 54 } 55 else 56 { 57 parent[r1]=temp; 58 parent[r2]=r1; 59 } 60 } 61 62 int kruskal(int x) 63 { 64 int sum=0; 65 int i; 66 for(i=1;i
             
              i) 92 { 93 point[cnt].u=i; 94 point[cnt].v=j; 95 point[cnt++].w=edge[i][j]; 96 } 97 } 98 } 99 sort(point+1,point+cnt,cmp);100 int q;101 scanf("%d",&q);102 for(i=1;i<=q;i++)103 {104 int u,v;105 scanf("%d%d",&u,&v);106 if(Find(u)!=Find(v))107 {108 Union(u,v);109 }110 }111 int ans=kruskal(cnt);112 printf("%d\n",ans);113 }114 return 0;115 }

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转载于:https://www.cnblogs.com/zafuacm/p/3199712.html

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