POJ 1251 + HDU 1301 Jungle Roads 【最小生成树】

题解

这是一道裸的最小生成树题，拿来练手，题目就不放了

个人理解  Prim有些类似最短路和贪心，不断找距当前点最小距离的点

Kruskal类似于并查集，不断找最小的边，如果不是一棵树的节点就合并为一颗树

AC代码：

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 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+;
const int MAXN = ;
const int maxn = ;
int n,m,v;
int pos,imin ;
int ans ;
char a,b ;
int vis[MAXN],dis[MAXN];
int mapp[MAXN][MAXN];
void Init(){
  mst(vis,);
  ans = ;
  for(int i =  ;i < n; i++)  dis[i] = inf;
  for(int i =  ;i < n; i++)
    for(int j = ; j < n; j++){
      if(i == j) mapp[i][j] = ;
      else mapp[i][j] = inf;
    }
}
void prim(){
  for(int i = ; i < n ; i++)
      dis[i]  = mapp[][i];
  dis[] = ;
  vis[] = ;
  for(int i =  ; i < n ; i ++) {
    pos = ;
    imin  = inf;
    for(int j =  ; j < n ; j++ )
        if(!vis[j]  && dis[j] < imin)   pos = j , imin = dis[j];
    vis[pos] = ;
    ans += imin ;
    for(int j = ; j < n; j++)
       if(!vis[j] && mapp[pos][j] < dis[j])   dis[j] = mapp[pos][j];
  }
}
int main(){
  while(cin >> n && n){
      Init();
      for(int i = ; i < n; i++){
        cin >> a >> m ;
        int st = a -'A';
        while(m--) {
          cin >> b  >> v ;
          int ed = b - 'A';
          mapp[st][ed] = v;
          mapp[ed][st] = v;
        }
      }
      prim();
      print(ans);
  }
  return ;
}

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+;
const int MAXN = ;
const int maxn = ;

struct node{
  int st,ed,v;
  bool operator < (node b) const{
     return v < b.v;
  }
}rod[MAXN];
int n,m;
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 = ;
  cnt = ;
  for(int i =  ; i < MAXN ; i++){
    pre[i] = i;
  }
}
void kruskal(){
  for(int i =  ;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(cin >> n && n){
      Init();
      char a,b ;
      int m,v;
      for(int i = ; i < n; i++){
        cin >> a >> m ;
        int st = a -'A';
        while(m--) {
          cin >> b  >> v ;
          int ed = b - 'A';
          rod[cnt].st = st;
          rod[cnt].ed = ed;
          rod[cnt++].v = v;
        }
      }
      sort(rod,rod+cnt);
      kruskal();
      print(ans);
  }
  return ;
}