无向图最小生成树（prim算法）

普里姆算法（Prim算法），图论中的一种算法，可在加权连通图里搜索最小生成树。意即由此算法搜索到的边子集所构成的树中，不但包括了连通图里的所有顶点，且其所有边的权值之和亦为最小。该算法于1930年由捷克数学家沃伊捷赫·亚尔尼克发现；并在1957年由美国计算机科学家罗伯特·普里姆独立发现；1959年，艾兹格·迪科斯彻再次发现了该算法。因此，在某些场合，普里姆算法又被称为DJP算法、亚尔尼克算法或普里姆－亚尔尼克算法

算法过程图解：遍历点，用贪心法选择与集合内的点相连的点的最小值；

模板：

 #include <iostream>
 #include <algorithm>
 #include <cmath>
 #include <cstdio>
 #include <string>
 #include <cstring>
 #include <vector>
 #include <queue>
 #include <stack>
  #include <set>
  #define INF 0x3f3f3f3f
  #define INFL 0x3f3f3f3f3f3f3f3f
  #define zero_(x,y) memset(x , y , sizeof(x))
  #define zero(x) memset(x , 0 , sizeof(x))
  #define MAX(x) memset(x , 0x3f ,sizeof(x))
  using namespace std;
  #define N 1005
  typedef long long LL ;
  int G[N][N];
  int minn[N];///寻找最小权；
  bool point[N];

  int main(){
      //freopen("in.txt","r",stdin);
      memset(point, , sizeof(point));
      int n, m, x, y, ans=;
      scanf("%d%d", &n, &m);
      for(int i=;i<=n;i++){
          for(int j = ; j <= n; j++){
              G[i][j] = INF;
          }
      }
      for(int i = ; i <= m; i++) {
          scanf("%d%d", &x, &y);
          scanf("%d", &G[x][y]);
          G[y][x] = G[x][y];
      }
      point[] = true;
      for(int i = ;i <= n; i++) minn[i] = G[][i];
      for(int i =  ; i <= n; i++){
          int k = , mi = INF;
          for(int j = ; j <= n; j++){
              if(!point[j] && minn[j] < mi){
                  mi = minn[j];
                  k = j;
              }
          }
          ans += minn[k];
          point[k] = true;
          for(int s = ;s <= n; s++){
              if(!point[s] && G[k][s] < minn[s])///保存了之前没有加入集合的边的权；
                  minn[s] = G[k][s];
          }
      }
      printf("%d\n", ans);
      return ;
  }

poj 1789

题意：点是7位小写字母，边是小写字母间的不同字母数量（1-7）。求最小生成树。

边权：the number of positions with different letters in truck type codes.

 #include <iostream>
 #include <algorithm>
 #include <stdlib.h>
 #include <time.h>
 #include <cmath>
 #include <cstdio>
 #include <string>
 #include <cstring>
 #include <vector>
 #include <queue>
 #include <stack>
 #include <set>

 #define c_false ios_base::sync_with_stdio(false); cin.tie(0)
 #define INF 0x3f3f3f3f
 #define INFL 0x3f3f3f3f3f3f3f3f
 #define zero_(x,y) memset(x , y , sizeof(x))
 #define zero(x) memset(x , 0 , sizeof(x))
 #define MAX(x) memset(x , 0x3f ,sizeof(x))
 #define swa(x,y) {LL s;s=x;x=y;y=s;}
 using namespace std ;
 #define N 2005

 const double PI = acos(-1.0);
 typedef long long LL ;
 int n, dis[N][N], minn[N];
 char x[];
 string type[N];
 bool point[N];

 int Distance(int i, int j){
     int len = , sum = ;
     for(int k = ; k < len; k++)
         if(type[i][k] != type[j][k])
             sum++;
     return sum;
 }
 int prim(){
     point[] = true;
     int ans = ;
     for(int i = ; i <= n; i++) minn[i] = dis[][i];
     for(int i = ; i <= n; i++){
         int k = ; int mi =INF;
         for(int j = ; j <= n; j++){
             if(!point[j] && minn[j] <mi){
                 mi = minn[j];
                 k = j;
             }
         }
         ans += minn[k];
         point[k] = true;
         for(int s = ; s <= n; s++){
             if(!point[s] && dis[k][s] <minn[s])
                 minn[s] = dis[k][s];
         }
     }
     return ans;
 }
 int main(){
     //freopen("in.txt","r",stdin);
     //freopen("out.txt","w",stdout);
     //ios_base::sync_with_stdio(false); cin.tie(0);
     while(~scanf("%d", &n) && n){
         zero(dis);zero(point);MAX(dis);
         for(int i = ; i <= n; i++){
             scanf("%s", x);
             type[i] = x;
         }
         for(int i = ; i<= n; i++){
             for(int j = ; j < i; j++){
                 dis[i][j] = dis[j][i] = Distance(i, j);
             }
         }
         printf("The highest possible quality is 1/%d.\n", prim());
     }
     return ;
 }

poj 2485

题意：求最小生成树的最大边权；(英语实在太烂)

point：minimize the length of the longest highway

 #include <iostream>
 #include <algorithm>
 #include <stdlib.h>
 #include <time.h>
 #include <cmath>
 #include <cstdio>
 #include <string>
 #include <cstring>
 #include <vector>
 #include <queue>
 #include <stack>
 #include <set>

 #define c_false ios_base::sync_with_stdio(false); cin.tie(0)
 #define INF 0x3f3f3f3f
 #define INFL 0x3f3f3f3f3f3f3f3f
 #define zero_(x,y) memset(x , y , sizeof(x))
 #define zero(x) memset(x , 0 , sizeof(x))
 #define MAX(x) memset(x , 0x3f ,sizeof(x))
 #define swa(x,y) {LL s;s=x;x=y;y=s;}
 using namespace std ;
 #define N 505

 const double PI = acos(-1.0);
 typedef long long LL ;
 int T, n, dis[N][N], minn[N];
 bool point[N];

 int prim(){
     point[] = true;
     int ans = ;
     for(int i = ; i <= n; i++) minn[i] = dis[][i];
     for(int i = ; i <= n; i++){
         int k = ; int mi =INF;
         for(int j = ; j <= n; j++){
             if(!point[j] && minn[j] <mi){
                 mi = minn[j];
                 k = j;
             }
         }
         if(ans < minn[k]) ans = minn[k];
         point[k] = true;
         for(int s = ; s <= n; s++){
             if(!point[s] && dis[k][s] <minn[s])
                 minn[s] = dis[k][s];
         }
     }
     return ans;
 }

 int main(){
     //freopen("in.txt","r",stdin);
     //freopen("out.txt","w",stdout);
     //ios_base::sync_with_stdio(false); cin.tie(0);
     scanf("%d", &T);
     while(T--){
         zero(point);zero(dis);zero(minn);
         scanf("%d", &n);
         for(int i = ; i <= n; i++){
             for(int j = ; j <= n; j++){
                 scanf("%d", &dis[i][j]);
             }
         }
         printf("%d\n", prim());
     }
     return ;
 }

poj 1258

简单模板题，同上；

 #include <iostream>
 #include <algorithm>
 #include <stdlib.h>
 #include <time.h>
 #include <cmath>
 #include <cstdio>
 #include <string>
 #include <cstring>
 #include <vector>
 #include <queue>
 #include <stack>
 #include <set>

 #define c_false ios_base::sync_with_stdio(false); cin.tie(0)
 #define INF 0x3f3f3f3f
 #define INFL 0x3f3f3f3f3f3f3f3f
 #define zero_(x,y) memset(x , y , sizeof(x))
 #define zero(x) memset(x , 0 , sizeof(x))
 #define MAX(x) memset(x , 0x3f ,sizeof(x))
 #define swa(x,y) {LL s;s=x;x=y;y=s;}
 using namespace std ;
 #define N 505

 const double PI = acos(-1.0);
 typedef long long LL ;
 int n, dis[N][N], minn[N];
 bool point[N];

 int prim(){
     point[] = true;
     int ans = ;
     zero(minn);zero(point);
     for(int i = ; i <= n; i++) minn[i] = dis[][i];
     for(int i = ; i <= n; i++){
         int k = ; int mi =INF;
         for(int j = ; j <= n; j++){
             if(!point[j] && minn[j] <mi){
                 mi = minn[j];
                 k = j;
             }
         }
         ans += minn[k];
         point[k] = true;
         for(int s = ; s <= n; s++){
             if(!point[s] && dis[k][s] <minn[s])
                 minn[s] = dis[k][s];
         }
     }
     return ans;
 }

 int main(){
     //freopen("in.txt","r",stdin);
     //freopen("out.txt","w",stdout);
     //ios_base::sync_with_stdio(false); cin.tie(0);
     while(~scanf("%d", &n) && n){
         zero(dis);
         for(int i = ; i <= n; i++){
             for(int j = ; j <= n; j++){
                 scanf("%d", &dis[i][j]);
             }
         }
         printf("%d\n", prim());
     }
     return ;
 }

poj 3026

题意：求最小生成树；

思路：prim + BFS

　　　这题的难点在于寻找邻接矩阵；

　　  用BFS 找到各点间的最小距离的点；

　然后就是套模板；