付翔的专栏 在鄙视中成长 记录成长的点滴

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3780    Accepted Submission(s): 1298

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.

#include<iostream>
#include
<string.h>
using namespace std;

#define infinity 123456789
#define max_vertexes 100

typedef
int Graph[max_vertexes][max_vertexes];

Graph G;
int total;
int lowcost[max_vertexes],closeset[max_vertexes],used[max_vertexes];
int father[max_vertexes];
void prim(Graph G,int vcount)
{

int i,j,k;

int min = infinity;

for (i=0;i<vcount;i++)
{
lowcost[i]
=G[0][i];
closeset[i]
=0
used[i]
=0;
father[i]
=-1
}
used[
0]=1

for (i=1;i<vcount;i++)
{
j
=0;

while (used[j]) j++;
min
= lowcost[j];

for (k=0;k<vcount;k++)

if ((!used[k])&&(lowcost[k]<min))
{
min
=lowcost[k];
j
=k;
}
father[j]
=closeset[j];
used[j]
=1;
total
+= min;

for (k=0;k<vcount;k++)

if (!used[k]&&(G[j][k]<lowcost[k]))
{
lowcost[k]
=G[j][k];
closeset[k]
=j;
}
}
}

int main()
{

int N,i,j,Q;

int x,y;

while(cin>>N)
{

total
= 0;

for(i =0; i< N;i++)
{

for(j = 0;j < N; j ++)
cin
>>G[i][j];
}
cin
>>Q;

for(i = 0; i < Q; i ++)
{
cin
>>x>>y;
G[x
-1][y-1= 0;
G[y
-1][x-1= 0;
}
prim(G,N);
cout
<< total<<endl;
}

return 0;
}

posted on 2010-09-02 23:48 付翔 阅读(269) 评论(0)  编辑 收藏 引用 所属分类: ACM 图论

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