HDU1233 还是畅通工程【最小生成树】

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.

 

Prim算法：

#include<iostream>
#include<cstdio>     //EOF,NULL
#include<cstring>    //memset
#include<cstdlib>    //rand,srand,system,itoa(int),atoi(char[]),atof(),malloc
#include<cmath>           //ceil,floor,exp,log(e),log10(10),hypot(sqrt(x^2+y^2)),cbrt(sqrt(x^2+y^2+z^2))
#include<algorithm>  //fill,reverse,next_permutation,__gcd,
#include<string>
#include<vector>
#include<queue>
#include<stack>
#include<utility>
#include<iterator>
#include<iomanip>             //setw(set_min_width),setfill(char),setprecision(n),fixed,
#include<functional>
#include<map>
#include<set>
#include<limits.h>     //INT_MAX
#include<bitset> // bitset<?> n
using namespace std;

typedef long long ll;
typedef pair<int,int> P;
#define all(x) x.begin(),x.end()
#define readc(x) scanf("%c",&x)
#define read(x) scanf("%d",&x)
#define read2(x,y) scanf("%d%d",&x,&y)
#define read3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define print(x) printf("%d\n",x)
#define mst(a,b) memset(a,b,sizeof(a))
#define disbit(x) x&-x
#define lson(x) x<<1
#define rson(x) x<<1|1
#define pb push_back
#define mp make_pair
const int INF =0x3f3f3f3f;
const int inf =0x3f3f3f3f;
const int mod = 1e9+7;
const int MAXN = 110;
const int maxn = 10010;
int n,m,v;
int pos,imin ;
int ans ;
int st,ed;
int vis[MAXN],dis[MAXN];
int mapp[MAXN][MAXN];
void Init(){
  mst(vis,0);
  ans = 0;
  for(int i = 1 ;i <= n; i++)  dis[i] = inf;
  for(int i = 1 ;i <= n; i++)
    for(int j = 1; j <= n; j++){
      if(i == j) mapp[i][j] = 0;
      else mapp[i][j] = inf;
    }
}
void prim(){
  for(int i = 1; i <= n ; i++)
      dis[i]  = mapp[1][i];
  dis[1] = 0;
  vis[1] = 1;
  for(int i = 1 ; i < n ; i ++) {
    pos = 1;
    imin  = inf;
    for(int j = 1 ; j <= n ; j++ )
        if(!vis[j]  && dis[j] < imin)  {
         pos = j , imin = dis[j];
        }
    vis[pos] = 1;
    ans += imin ;
    for(int j = 1; j <= n; j++)
       if(!vis[j] && mapp[pos][j] < dis[j])   dis[j] = mapp[pos][j];
    }
}
int main(){
  while(read(n) && n){
      m =  n * (n - 1) / 2;
      Init();
      for(int i = 0; i < m; i++){
          read3(st,ed,v);
          mapp[st][ed] = v;
          mapp[ed][st] = v;
      }
      prim();
      print(ans);
  }
  return 0;
}

 

Kruskal算法：

#include<iostream>
#include<cstdio>     //EOF,NULL
#include<cstring>    //memset
#include<cstdlib>    //rand,srand,system,itoa(int),atoi(char[]),atof(),malloc
#include<cmath>           //ceil,floor,exp,log(e),log10(10),hypot(sqrt(x^2+y^2)),cbrt(sqrt(x^2+y^2+z^2))
#include<algorithm>  //fill,reverse,next_permutation,__gcd,
#include<string>
#include<vector>
#include<queue>
#include<stack>
#include<utility>
#include<iterator>
#include<iomanip>             //setw(set_min_width),setfill(char),setprecision(n),fixed,
#include<functional>
#include<map>
#include<set>
#include<limits.h>     //INT_MAX
#include<bitset> // bitset<?> n
using namespace std;

typedef long long ll;
typedef pair<int,int> P;
#define all(x) x.begin(),x.end()
#define readc(x) scanf("%c",&x)
#define read(x) scanf("%d",&x)
#define read2(x,y) scanf("%d%d",&x,&y)
#define read3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define print(x) printf("%d\n",x)
#define mst(a,b) memset(a,b,sizeof(a))
#define lowbit(x) x&-x
#define lson(x) x<<1
#define rson(x) x<<1|1
#define pb push_back
#define mp make_pair
const int INF =0x3f3f3f3f;
const int inf =0x3f3f3f3f;
const int mod = 1e9+7;
const int MAXN = 10005;
const int maxn = 10010;
struct node{
  int st,ed,v;
  bool operator < (node b) const{
     return v < b.v;
  }
}rod[MAXN];
int n,m,v;
int cnt,ans;
int pre[MAXN];

int find(int x){ return x == pre[x] ? x : pre[x] = find(pre[x]);}
bool join(int x,int y){
    if(find(x)!=find(y)){
      pre[find(y)] = find(x);
      return true;
    }
    return false;
}
void Init(){
  ans = 0;
  cnt = 0;
  for(int i = 0 ; i < MAXN ; i++){
    pre[i] = i;
  }
}
void kruskal(){
  for(int i = 0 ;i < cnt ; i++){
    int mp1 = find(rod[i].st);
    int mp2 = find(rod[i].ed);
    if(join(mp1,mp2)) ans+= rod[i].v;
  }
}
int main(){
  while(read(n) && n){
        Init();
        int a,b;
        for(int i = 0; i< n * (n - 1) / 2;i++){
          read3(a,b,v);
          rod[cnt].st = a;
          rod[cnt].ed = b;
          rod[cnt++].v = v;
        }
      sort(rod,rod+cnt);
      kruskal();
      print(ans);
  }

}