hdu1102Constructing Roads

Constructing Roads

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12166    Accepted Submission(s): 4621

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

3 0 990 692 990 0 179 692 179 0 1 1 2

 

Sample Output

179

prim算法适用于稠密图

kruskal算法适用于稀疏图

prim：

#include
#include
#include
#include
#include
#include
using namespace std;
const int LEN=105;
int path[LEN][LEN], vis[LEN], dis[LEN];
void prim(int& cost, const int& n)
{
    int i, j, site;
    for(i=1; i<=n; i++)
        dis[i]=path[i][1], vis[i]=0;
    vis[1]=1;
    dis[1]=0;
    cost=0;
    for(i=2; i<=n; i++)
    {
        site=-1;
        for(j=1; j<=n; j++)
        {
            if( vis[j] ) continue;
            if( site==-1 || dis[j]path[j][site] )
                dis[j]=path[j][site];
        }
    }
}
int main()
{
    int N, i, j, a, b, Q;
    while( cin>>N )
    {
        int cost=0;
        for(i=1; i<=N; i++)
            for(j=1; j<=N; j++)
            {
                scanf("%d", &path[i][j]);
            }
        scanf("%d", &Q);
        for(i=0; i

kruskal：

#include
#include
#include
#include
#include
using namespace std;
const int LEN=105;
int path[LEN][LEN], vis[LEN], dis[LEN];
struct node
{
    int st, ed, w;
}road[LEN*LEN];
int find_fa(const int& a)
{
    return vis[a]==a ? a : vis[a]=find_fa(vis[a]);
}
void unite(const int &x, const int &y)
{
    if( x==y )
        return ;
    if( dis[x]>dis[y] )
        vis[y]=x;
    else
    {
        vis[x]=y;
        if( dis[x]==dis[y] )
            dis[y]++;
    }
}
int cmp(const void *a, const void *b)
{
    node x=*(node *)a;
    node y=*(node *)b;
    return x.w-y.w;
}
int kruskal(const int& n)
{
    int cost=0, i, j, k=0;
    for(i=1; i<=n; i++)
        for(j=1; j<=n; j++)
        {
            road[k].st=i;
            road[k].ed=j;
            road[k++].w=path[i][j];
        }
    qsort(road, k, sizeof(road[1]), cmp);
    for(i=1; i<=n; i++)
    {
        vis[i]=i;
        dis[i]=0;
    }
    k=0;
    for(j=0; j>n )
    {
        for(i=1; i<=n; i++)
            for(j=1; j<=n; j++)
                scanf("%d", &path[i][j]);
        scanf("%d", &Q);
        for(i=0; i