Constructing Roads

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

Total Submission(s): 1914 Accepted Submission(s): 628

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

30 990 692990 0 179692 179 011 2

Sample Output

解题：

还是用克鲁斯卡尔算法。题目大意是：有N个村庄，给定我们每个村庄之间的距离，而两个村庄可以间接相连。然后告诉我们原本有几个村庄已经相连有路相通。要让我们求还要建多长的路，最小的。可以用最小生成树，先把给定的已相连的村庄之间的权值赋为0，其余跟求最小生成树一样。

#include <iostream>using namespace std;#define MaxSize 101#define INF 0x7ffffffftypedef struct node   {       int u;      //边的起始顶点       int v;      //边的终止顶点       int w;      //边的权值   }Edge;     int cmp( const void *a , const void *b )        {            struct node *c = (node *)a;           struct node *d = (node *)b;           if(c->w != d->w) return c->w - d->w;           else return d->u - c->u;       }      void kruskal(int dist[][MaxSize],int n,int& sum)   {       int i,j,u1,v1,sn1,sn2,k;       int vset[MaxSize];         //辅助数组，判定两个顶点是否连通       Edge E[MaxSize*MaxSize];             //存放所有的边       k=0;                            //E数组的下标从0开始       for (i=0;i<n;i++)       {           for (j=0;j<n;j++)           {               if (dist[i][j]!=INF)               {                   E[k].u=i;                   E[k].v=j;                   E[k].w=dist[i][j];                   k++;               }           }       }          qsort(E,k,sizeof(E[0]),cmp);        //快速排序，按权值从小到大排列           for (i=0;i<n;i++)                  //初始化辅助数组       {           vset[i]=i;       }       k=1;                //生成的边数，最后要刚好为总边数       j=0;                //E中的下标       while (k<n)       {           u1=E[j].u;           v1=E[j].v;           sn1=vset[u1];           sn2=vset[v1];   //得到两顶点属于的集合编号           if (sn1!=sn2)   //不在同一集合编号内的话，把边加入最小生成树           {                   			sum+=E[j].w;            k++;               for (i=0;i<n;i++)               {                   if (vset[i]==sn2)                   {                       vset[i]=sn1;                   }               }                      }           j++;       }   }  int main(){		int i,j,n,q,sum,dist[MaxSize][MaxSize];	while (cin>>n)	{		sum=0;		for(i=0;i<n;i++)			for(j=0;j<n;j++)				cin>>dist[i][j];				cin>>q;		while(q--)		{			cin>>i>>j;			dist[i-1][j-1]=0;			dist[j-1][i-1]=0;		}		kruskal(dist,n,sum);		cout<<sum<<endl;	}	return 0;}

转载地址：http://ucid.baihongyu.com/

你可能感兴趣的文章

MySQL中ON DUPLICATE KEY UPDATE的介绍与使用、批量更新、存在即更新不存在则插入