Jungle Roads

最小生成树的两种算法
Prim；

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int map[121][121];
int book[121];
int dis[121];
int n,inf=9999999;
void Prim()
{
    int i,j,u;
    memset(book, 0, sizeof(book));
    //printf("%d\n",n);
    for(int i=1; i<=n; i++)
    {
        dis[i] = map[1][i];
       // printf("%d\n",dis[i]);
    }
    book[1] = 1;
    int sum= 0;
    for(int i=1; i<=n; i++)
    {
        int minn = inf;
        for(int j=0; j<=n; j++)
        {
            if(dis[j]<minn&&!book[j])
            {
                minn = dis[j];
                u = j;
            }
        }
        sum += minn;
        book[u] = 1;
        for(int j=0; j<=n; j++)
        {
            if(map[u][j]<dis[j]&&book[j]==0)
            {
                dis[j] = map[u][j];
            }
        }
    }
    printf("%d\n", sum);
}
int main()
{
    int a, b, f, e;
    char c, d;
    while(~scanf("%d", &n)&&n)
    {
        memset(map,inf,sizeof(map));
        for(int i=1;i<n;i++)
        {
            scanf("\n%c",&c);
            a=c-'A'+1;
            scanf("%d",&f);
            for(int j=0;j<f;j++)
            {
                scanf("\n%c %d",&d,&e);
                b=d-'A'+1;
                map[a][b]=map[b][a]=e;
            }
        }
        Prim();
    }
    return 0;
}

kruskal

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int n,dot,inf=9999999;
int pre[150];
struct node
{
    int u;
    int v;
    int cost;
}p[5002];
int find(int x)
{
    if(pre[x]==x)
        return x;
    else
    {
        pre[x]=find(pre[x]);
        return pre[x];
    }
}
bool cmp(node x,node y)
{
    return x.cost<y.cost;
}
int main()
{
    int i,j;
    char a,b,c[150],d;
    while(~scanf("%d",&n)&&n)
    {
        for(i=0;i<n;i++)
        {
            pre[i]=i;
        }
        int size=0,t,co;
        for(i=1;i<n;i++)
        {
            scanf("\n%c %d",&a,&t);
            for(j=0;j<t;j++)
            {
                scanf("\n%c %d",&b,&co);
                p[size].u=a-'A';
                p[size].v=b-'A';
                p[size].cost=co;
                size++;
            }
        }
        sort(p,p+size,cmp);
        int ans=0;
        for(i=0;i<size;i++)
        {
            int r1=find(p[i].u);
            int r2=find(p[i].v);
            if(r1!=r2)
            {
                pre[r1]=r2;
                ans+=p[i].cost;
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}

来源：CSDN

作者：RUBGH

链接：https://blog.csdn.net/RUBGH/article/details/104682671