ACM/ICPC 之 四道MST-Prim解法(POJ1258-POJ1751-POJ2349-POJ3026)

　　四道MST，适合Prim解法，也可以作为MST练习题。

　　题意包括在代码中。

---

POJ1258-Agri Net

　　水题

 //Prim-没什么好说的
 //接受一个邻接矩阵，求MST
 //Time:0Ms    Memory:220K
 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<algorithm>
 using namespace std;
 #define MAX 105
 #define INF 0x3f3f3f3f
 int n, m;
 int d[MAX][MAX];
 int lowcost[MAX];
 bool v[MAX];
 void prim()
 {
     int minroad = ;
     memset(v, false, sizeof(v));
     v[] = true;
     for (int i = ; i < n; i++)
         lowcost[i] = d[i][];
     for (int i = ; i < n; i++)
     {
         double mind = INF;
         int k;
         for (int j = ; j < n; j++)
         {
             if (!v[j] && mind > lowcost[j])
             {
                 mind = lowcost[j];
                 k = j;
             }
         }

         minroad += lowcost[k];
         v[k] = true;
         for (int j = ; j < n; j++)
             if (!v[j])    lowcost[j] = min(d[k][j], lowcost[j]);
     }
     printf("%d\n", minroad);
 }
 int main()
 {
     while (scanf("%d", &n) != EOF)
     {
         for (int i = ; i < n; i++)
             for (int j = ; j < n; j++)
                 scanf("%d", &d[i][j]);
         prim();
     }
     return ;
 }

---

POJ1751(ZOJ2048)-Highways

 //Prim-好题
 //ZOJ2048-POJ1751
 //ZOJ中多组样例，两个样例间有一个空格(否则会WA)
 //需要记录上一个节点-注意内存限制在10^4K内
 //有M个城市已经有通路，输出让N个城市生成最短通路的各边，Special Judge-输出次序不定
 //Time:94Ms    Memory:4648K
 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<cmath>
 #include<algorithm>
 using namespace std;

 #define MAX 755
 #define INF 0x3f3f3f3f
 #define POW2(x) ((x)*(x))
 #define DIS(i,j) (sqrt(POW2(p[i][0] - p[j][0]) + POW2(p[i][1] - p[j][1])))

 int n, m;
 double p[MAX][];
 int fa[MAX];    //记录上一个顶点
 double d[MAX][MAX];
 double lowcost[MAX];
 bool v[MAX];

 void prim()
 {
     memset(v, false, sizeof(v));
     v[] = true;
     for (int i = ; i <= n; i++)
     {
         lowcost[i] = d[i][];
         fa[i] = ;
     }
     for (int i = ; i <= n; i++)
     {
         double mind = INF;
         int k;
         for (int j = ; j <= n; j++)
         {
             if (!v[j] && mind > lowcost[j])
             {
                 mind = lowcost[j];
                 k = j;
             }
         }
         if (mind > 1e-)
             printf("%d %d\n", fa[k], k);

         v[k] = true;
         for (int j = ; j <= n; j++)
         {
             if (!v[j] && lowcost[j] > d[k][j])
             {
                 lowcost[j] = d[k][j];
                 fa[j] = k;
             }
         }
     }
 }

 int main()
 {
     scanf("%d", &n);
     for (int i = ; i <= n; i++)
     {
         scanf("%lf%lf", &p[i][], &p[i][]);
         for (int j = ; j < i; j++)
             d[i][j] = d[j][i] = DIS(i, j);
     }
     scanf("%d", &m);
     for (int i = ; i < m; i++)
     {
         int v1, v2;
         scanf("%d%d", &v1, &v2);
         d[v1][v2] = d[v2][v1] = ;
     }
     prim();
     return ;
 }

---

POJ2349(ZOJ1914)-Arctic Network

 //Prim
 //POJ2349-ZOJ1914
 //有n个前哨可以通过卫星通信(无距离限制)，总共m个前哨，相互通信可以通过无线电通信(有距离限制),求所需无线电信号最短距离
 //定理：如果去掉所有权值大于d的边后，最小生成树被分割成为k个连通支，图也被分割成为k个连通支(可尝试证明)
 //Time:47Ms    Memory:2164K
 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<cmath>
 #include<algorithm>
 using namespace std;

 #define MAX 501
 #define INF 0x3f3f3f3f
 #define POW2(x) ((x)*(x))
 #define DIS(i,j) (sqrt(POW2(p[i][0] - p[j][0]) + POW2(p[i][1] - p[j][1])))

 int n, m;
 double p[MAX][];    //point
 double d[MAX][MAX];    //distance
 double lowcost[MAX];
 bool v[MAX];

 void prim()
 {
     memset(lowcost, , sizeof(lowcost));
     memset(v, false, sizeof(v));
     v[] = true;
     for (int i = ; i < m; i++)
         lowcost[i] = d[i][];
     for (int i = ; i < m; i++)
     {
         int mind = INF;
         int k;
         for (int j = ; j < m; j++)
         {
             if (!v[j] && mind > lowcost[j])
             {
                 mind = lowcost[j];
                 k = j;
             }
         }
         v[k] = true;
         for (int j = ; j < m; j++)
             if(!v[j]) lowcost[j] = min(d[k][j], lowcost[j]);
     }
 }

 int main()
 {
     int T;
     scanf("%d", &T);
     while (T--)
     {
         scanf("%d%d", &n, &m);
         for (int i = ; i < m; i++)
         {
             scanf("%lf%lf", &p[i][], &p[i][]);
             for (int j = ; j < i; j++)
                 d[i][j] = d[j][i] = DIS(i, j);
         }

         prim();
         sort(lowcost, lowcost + m);
         printf("%.2lf\n", lowcost[m - n]);
         //G++需要使用printf("%.2f\n", lowcost[m-n]);
         //原因查了半天，好像是因为新版GCC标准中将%f和%lf合并为%f的意思
     }
     return ;
 }

---

POJ3026-Borg Maze

 //Prim+BFS
 //总是心想着要创造一个新算法，结果越想越麻烦...
 //保险做法:找出每个点间的距离，再进行Prim
 //Time:79Ms    Memory:292K
 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<queue>
 #include<algorithm>
 using namespace std;

 #define MAX 125        //MAX 50的话会RE或WA(博主在55RE，在100WA)

 struct Point{
     int x, y;
     int step;
 }p;

 int r, c;
 char board[MAX][MAX];
 int d[MAX][MAX];
 int num[MAX][MAX], cnt;
 int lowcost[MAX];
 bool v[MAX][MAX];
 int mov[][] = { {,}, {-,}, {,}, {,-} };

 void bfs(Point p)
 {
     memset(v, false, sizeof(v));
     v[p.x][p.y] = true;
     int np = num[p.x][p.y];
     queue<Point> q;
     p.step = ;
     q.push(p);
     while (!q.empty())
     {
         Point cur = q.front();
         q.pop();
         for (int i = ; i < ; i++)
         {
             Point t = cur;
             t.x += mov[i][];
             t.y += mov[i][];
             if (t.x >  && t.y >  && t.x < r && t.y < c && !v[t.x][t.y])
             {
                 if (board[t.x][t.y] == '#')    continue;
                 int nt = num[t.x][t.y];
                 t.step++;
                 v[t.x][t.y] = true;
                 if (board[t.x][t.y] == 'A' || board[t.x][t.y] == 'S')
                     d[nt][np] = t.step;
                 q.push(t);
             }
         }
     }
 }

 void prim()
 {
     int v[MAX];
     memset(v, false, sizeof(v));
     memset(lowcost, 0x3f, sizeof(lowcost));
     v[] = true;
     for (int i = ; i < cnt; i++)
         lowcost[i] = d[i][];

     int minv = ;
     for (int i = ; i < cnt; i++)
     {
         int mind = 0x3f3f3f3f;
         int k;
         for (int j = ; j < cnt; j++)
         {
             if (!v[j] && mind > lowcost[j])
             {
                 mind = lowcost[j];
                 k = j;
             }
         }
         minv += mind;
         v[k] = true;
         for (int j = ; j < cnt; j++)
             if (!v[j]) lowcost[j] = min(lowcost[j], d[k][j]);
     }
     printf("%d\n", minv);
 }

 int main()
 {
     int T;
     scanf("%d", &T);
     while (T--)
     {
         cnt = ;
         scanf("%d%d", &c, &r);
         gets_s(board[]);
         for (int i = ; i < r; i++)
         {
             gets_s(board[i], MAX);
             for (int j = ; j < c; j++)
                 if (board[i][j] == 'S' || board[i][j] == 'A')
                     num[i][j] = cnt++;
         }
         for (int i = ; i < r; i++)
             for (int j = ; j < c;j++)
                 if (board[i][j] == 'S' || board[i][j] == 'A')
                 {
                     p.x = i; p.y = j;
                     bfs(p);
                 }
         prim();
     }
     return ;
 }