还是畅通工程 HDU1233(最小生成树)

还是畅通工程
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 29254 Accepted Submission(s): 13088


Problem Description
某省调查乡村交通状况,得到的统计表中列出了任意两村庄间的距离。省政府“畅通工程”的目标是使全省任何两个村庄间都可以实现公路交通(但不一定有直接的公路相连,只要能间接通过公路可达即可),并要求铺设的公路总长度为最小。请计算最小的公路总长度。


Input
测试输入包含若干测试用例。每个测试用例的第1行给出村庄数目N ( < 100 );随后的N(N-1)/2行对应村庄间的距离,每行给出一对正整数,分别是两个村庄的编号,以及此两村庄间的距离。为简单起见,村庄从1到N编号。
当N为0时,输入结束,该用例不被处理。


Output
对每个测试用例,在1行里输出最小的公路总长度。


Sample Input

3
1 2 1
1 3 2
2 3 4
4
1 2 1
1 3 4
1 4 1
2 3 3
2 4 2
3 4 5
0



Sample Output

3
5

Hint
Hint

Huge input, scanf is recommended.



Source

浙大计算机研究生复试上机考试-2006年


prim 算法

//#include<bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<string>
#include<queue>
#include<vector>
#include<map>
#include<cstdlib>
#include<set>
#include<stack>
#include<cstring>
using namespace std;
template<class T>inline T read(T&x)
{
    char c;
    while((c=getchar())<=32)if(c==EOF)return -1;
    bool ok=false;
    if(c=='-')ok=true,c=getchar();
    for(x=0; c>32; c=getchar())
        x=x*10+c-'0';
    if(ok)x=-x;
    return 1;
}
template<class T> inline T read_(T&x,T&y)
{
    return read(x)!=-1&&read(y)!=-1;
}
template<class T> inline T read__(T&x,T&y,T&z)
{
    return read(x)!=-1&&read(y)!=-1&&read(z)!=-1;
}
template<class T> inline void write(T x)
{
    if(x<0)putchar('-'),x=-x;
    if(x<10)putchar(x+'0');
    else write(x/10),putchar(x%10+'0');
}
template<class T>inline void writeln(T x)
{
    write(x);
    putchar('\n');
}
//-------ZCC IO template------
const int maxn=101;
const double inf=999999999;
#define lson (rt<<1),L,M
#define rson (rt<<1|1),M+1,R
#define M ((L+R)>>1)
#define For(i,t,n) for(int i=(t);i<(n);i++)
typedef long long  LL;
typedef double DB;
typedef pair<int,int> P;
#define bug printf("---\n");
#define mod 10007

int G[maxn][maxn];
int vis[maxn];
int mincost[maxn];
int V,E;
void prim()
{
    For(i,0,V+1)
    {
        mincost[i]=inf;
        vis[i]=false;
    }
    mincost[1]=0;
    int ans=0;
    while(true)
    {
        int v=-1;
        For(u,1,V+1)if(!vis[u]&&(v==-1||mincost[v]>mincost[u]))v=u;
        if(v==-1)break;
        vis[v]=true;
        ans+=mincost[v];
        For(u,1,V+1)mincost[u]=min(mincost[u],G[v][u]);
    }
    writeln(ans);
}
int main()
{
  //#ifndef ONLINE_JUDGE
   //freopen("in.txt","r",stdin);
 // #endif // ONLINE_JUDGE
    int n,m,i,j,k,t;
    while(read(V)&&V)
    {
        E=V*(V-1)/2;
        while(E--)
        {
            int a,b,c;
            read__(a,b,c);
            G[a][b]=G[b][a]=c;
        }
        prim();
    }
    return 0;
}

Kruskal算法

//#include<bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<string>
#include<queue>
#include<vector>
#include<map>
#include<cstdlib>
#include<set>
#include<stack>
#include<cstring>
using namespace std;
template<class T>inline T read(T&x)
{
    char c;
    while((c=getchar())<=32)if(c==EOF)return -1;
    bool ok=false;
    if(c=='-')ok=true,c=getchar();
    for(x=0; c>32; c=getchar())
        x=x*10+c-'0';
    if(ok)x=-x;
    return 1;
}
template<class T> inline T read_(T&x,T&y)
{
    return read(x)!=-1&&read(y)!=-1;
}
template<class T> inline T read__(T&x,T&y,T&z)
{
    return read(x)!=-1&&read(y)!=-1&&read(z)!=-1;
}
template<class T> inline void write(T x)
{
    if(x<0)putchar('-'),x=-x;
    if(x<10)putchar(x+'0');
    else write(x/10),putchar(x%10+'0');
}
template<class T>inline void writeln(T x)
{
    write(x);
    putchar('\n');
}
//-------ZCC IO template------
const int maxn=101;
const double inf=999999999;
#define lson (rt<<1),L,M
#define rson (rt<<1|1),M+1,R
#define M ((L+R)>>1)
#define For(i,t,n) for(int i=(t);i<(n);i++)
typedef long long  LL;
typedef double DB;
typedef pair<int,int> P;
#define bug printf("---\n");
#define mod 10007
struct edge
{
    int from,to,cost;
    edge(){}
    edge(int a,int b,int c){from=a,to=b,cost=c;}
    bool operator<(const edge&tmp)const{
    return cost<tmp.cost;
    }
}es[maxn*maxn];

int f[maxn];
void init(int n)
{
    For(i,0,n+1)f[i]=i;
}
int find(int x)
{
    return x==f[x]?x:(f[x]=find(f[x]));
}
void unio(int x,int y)
{
    x=find(x);y=find(y);
    if(x==y)return ;
    f[x]=y;
}
bool same(int x,int y)
{
    return find(x)==find(y);
}
int main()
{
  //#ifndef ONLINE_JUDGE
   //freopen("in.txt","r",stdin);
 // #endif // ONLINE_JUDGE
    int n,m,i,j,k,t;
    int V,E;
    while(read(V)&&V)
    {
        E=V*(V-1)/2;
        for(i=0;i<E;i++)
        {
            int a,b,c;
            read__(a,b,c);
            es[i]=edge(a,b,c);
        }
        init(V);
        sort(es,es+E);
        int ans=0;
        for(i=0;i<E;i++)
        {
            edge e=es[i];
            if(!same(e.to,e.from))
            {
                ans+=e.cost;
                unio(e.to,e.from);
            }
        }
        writeln(ans);
    }
    return 0;
}

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