求最小生成树(Prim算法)(1075)

Description

     求出给定无向带权图的最小生成树。图的定点为字符型,权值为不超过100的整形。在提示中已经给出了部分代码,你只需要完善Prim算法即可。

Input

     第一行为图的顶点个数n
     第二行为图的边的条数e
     接着e行为依附于一条边的两个顶点和边上的权值

Output

     最小生成树中的边。

Sample Input

ABCDEF
A B 6
A C 1
A D 5
B C 5
C D 5
B E 3
E C 6
C F 4
F D 2
E F 6
 
Sample Output
 
(A,C)(C,F)(F,D)(C,B)(B,E)
 
 
#include<iostream>
#include<cstring>
#include<cstdlib>
using namespace std;
   
typedef struct
{
    int n;
    int e;
    char data[500];
    int edge[500][500];
}Graph;
   
typedef struct
{
    int index;
    int cost;
}mincost;
   
typedef struct
{
    int x;
    int y;
    int weight;
}EDGE;
   
   
typedef struct
{
    int index;
    int flag;
}F;
   
void create(Graph &G, int n, int e)
{
    int i, j, k, w;
    char a, b;
    for (i = 0; i< n; i++)
        cin >> G.data[i];
    for (i = 0; i< n; i++)
    for (j = 0; j< n; j++)
    {
        if (i == j)
            G.edge[i][j] = 0;
        else
            G.edge[i][j] = 100;
    }
   
    for (k = 0; k< e; k++)
    {
        cin >> a;
        cin >> b;
        cin >> w;
        for (i = 0; i< n; i++)
        if (G.data[i] == a) break;
        for (j = 0; j< n; j++)
        if (G.data[j] == b) break;
   
        G.edge[i][j] = w;
        G.edge[j][i] = w;
    }
    G.n = n;
    G.e = e;
}
#define inf 32767
void Prim(Graph &G, int v)
{
    int lowcost[100];
    int min, closest[100], i, j, k;
    for (i = 0; i < G.n; i++)
    {
        lowcost[i] = G.edge[v][i];
        closest[i] = v;
    }
    for (i = 1; i < G.n; i++)
    {
        min = inf;
        for (j = 0; j < G.n; j++)
        {
            if (lowcost[j] && lowcost[j] < min)
            {
                min = lowcost[j];
                k = j;
            }
        }
        cout << '(' << G.data[closest[k]] << ',' << G.data[k] << ')';
        lowcost[k] = 0;
        for (j = 0; j < G.n; j++)
            if (G.edge[k][j] && G.edge[k][j] < lowcost[j])
            {
                lowcost[j] = G.edge[k][j];
                closest[j] = k;
        }
    }
}
   
   
   
int main()
{
    Graph my;
    int n, e;
    cin >> n >> e;
    create(my, n, e);
    Prim(my, 0);
    return 0;
}
View Code

 

 
posted @ 2015-06-16 09:10  soldierwing  阅读(422)  评论(3编辑  收藏  举报