POJ-1287-Networking（最小生成树 普利姆）

B - Networking
Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u
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Status

Practice

POJ 1287
Description
You are assigned to design network connections between certain points in a wide area. You are given a set of points in the area, and a set of possible routes for the cables that may connect pairs of points. For each possible route between two points, you are given the length of the cable that is needed to connect the points over that route. Note that there may exist many possible routes between two given points. It is assumed that the given possible routes connect (directly or indirectly) each two points in the area.
Your task is to design the network for the area, so that there is a connection (direct or indirect) between every two points (i.e., all the points are interconnected, but not necessarily by a direct cable), and that the total length of the used cable is minimal.
Input
The input file consists of a number of data sets. Each data set defines one required network. The first line of the set contains two integers: the first defines the number P of the given points, and the second the number R of given routes between the points. The following R lines define the given routes between the points, each giving three integer numbers: the first two numbers identify the points, and the third gives the length of the route. The numbers are separated with white spaces. A data set giving only one number P=0 denotes the end of the input. The data sets are separated with an empty line.
The maximal number of points is 50. The maximal length of a given route is 100. The number of possible routes is unlimited. The nodes are identified with integers between 1 and P (inclusive). The routes between two points i and j may be given as i j or as j i.
Output
For each data set, print one number on a separate line that gives the total length of the cable used for the entire designed network.
Sample Input
1 0

2 3
1 2 37
2 1 17
1 2 68

3 7
1 2 19
2 3 11
3 1 7
1 3 5
2 3 89
3 1 91
1 2 32

5 7
1 2 5
2 3 7
2 4 8
4 5 11
3 5 10
1 5 6
4 2 12

0
Sample Output
0
17
16
26

稠密图，普利姆的模板。

代码

#include<stdio.h>#include<iostream>#include<algorithm>#include<string.h>#include<math.h>using namespace std;//最小生成树//普利姆算法const int maxn=55;const int INF=99999999;int map[maxn][maxn];//邻接矩阵存图int dis[maxn];//点到生成树的最小距离int vis[maxn];//生成树外的点标记为0int N;//点的数量int M;//边的数量void init()//初始化邻接矩阵并接收数据{    for(int i=1; i<=N; i++)        for(int j=1; j<=N; j++)            i==j?map[i][j]=0:map[i][j]=INF;    for(int i=0; i<M; i++)    {        int u,v,w;        scanf("%d%d%d",&u,&v,&w);        if(map[u][v]>w)        {            map[u][v]=w;            map[v][u]=w;//无向图        }    }    memset(vis,0,sizeof(vis));//初始化所有点都在生成树以外    vis[1]=1;//把1放入生成树    for(int i=1; i<=N; i++)        dis[i]=map[i][1];//等待松弛}void Prim()//普利姆算法{    int min_dis=0;//记录最短路径    for(int i=1; i<N; i++)    {        int minn=INF;        int point_minn;//记录最小值的下标        for(int j=1; j<=N; j++)            if(vis[j]==0&&minn>dis[j])            {                minn=dis[j];                point_minn=j;            }        if(minn==INF)            break;        vis[point_minn]=1;        min_dis+=dis[point_minn];        for(int k=1; k<=N; k++)            if(vis[k]==0&&dis[k]>map[point_minn][k])                dis[k]=map[point_minn][k];    }    printf("%d\n",min_dis);}int main(){    while(~scanf("%d",&N)&&N)    {        scanf("%d",&M);        if(M==0)        {            printf("0\n");            continue;        }        init();        Prim();    }    return 0;}