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
1 //Prim-没什么好说的
2 //接受一个邻接矩阵，求MST
3 //Time:0Ms Memory:220K
4 #include<iostream>
5 #include<cstring>
6 #include<cstdio>
7 #include<algorithm>
8 using namespace std;
9 #define MAX 105
10 #define INF 0x3f3f3f3f
11 int n, m;
12 int d[MAX][MAX];
13 int lowcost[MAX];
14 bool v[MAX];
15 void prim()
16 {
17 int minroad = 0;
18 memset(v, false, sizeof(v));
19 v[0] = true;
20 for (int i = 1; i < n; i++)
21 lowcost[i] = d[i][0];
22 for (int i = 1; i < n; i++)
23 {
24 double mind = INF;
25 int k;
26 for (int j = 1; j < n; j++)
27 {
28 if (!v[j] && mind > lowcost[j])
29 {
30 mind = lowcost[j];
31 k = j;
32 }
33 }
34
35 minroad += lowcost[k];
36 v[k] = true;
37 for (int j = 1; j < n; j++)
38 if (!v[j]) lowcost[j] = min(d[k][j], lowcost[j]);
39 }
40 printf("%d\n", minroad);
41 }
42 int main()
43 {
44 while (scanf("%d", &n) != EOF)
45 {
46 for (int i = 0; i < n; i++)
47 for (int j = 0; j < n; j++)
48 scanf("%d", &d[i][j]);
49 prim();
50 }
51 return 0;
52 }
POJ1751(ZOJ2048)-Highways
1 //Prim-好题
2 //ZOJ2048-POJ1751
3 //ZOJ中多组样例，两个样例间有一个空格(否则会WA)
4 //需要记录上一个节点-注意内存限制在10^4K内
5 //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
6 //Time:94Ms Memory:4648K
7 #include<iostream>
8 #include<cstring>
9 #include<cstdio>
10 #include<cmath>
11 #include<algorithm>
12 using namespace std;
13
14 #define MAX 755
15 #define INF 0x3f3f3f3f
16 #define POW2(x) ((x)*(x))
17 #define DIS(i,j) (sqrt(POW2(p[i][0] - p[j][0]) + POW2(p[i][1] - p[j][1])))
18
19 int n, m;
20 double p[MAX][2];
21 int fa[MAX]; //记录上一个顶点
22 double d[MAX][MAX];
23 double lowcost[MAX];
24 bool v[MAX];
25
26 void prim()
27 {
28 memset(v, false, sizeof(v));
29 v[1] = true;
30 for (int i = 2; i <= n; i++)
31 {
32 lowcost[i] = d[i][1];
33 fa[i] = 1;
34 }
35 for (int i = 2; i <= n; i++)
36 {
37 double mind = INF;
38 int k;
39 for (int j = 2; j <= n; j++)
40 {
41 if (!v[j] && mind > lowcost[j])
42 {
43 mind = lowcost[j];
44 k = j;
45 }
46 }
47 if (mind > 1e-5)
48 printf("%d %d\n", fa[k], k);
49
50 v[k] = true;
51 for (int j = 2; j <= n; j++)
52 {
53 if (!v[j] && lowcost[j] > d[k][j])
54 {
55 lowcost[j] = d[k][j];
56 fa[j] = k;
57 }
58 }
59 }
60 }
61
62 int main()
63 {
64 scanf("%d", &n);
65 for (int i = 1; i <= n; i++)
66 {
67 scanf("%lf%lf", &p[i][0], &p[i][1]);
68 for (int j = 1; j < i; j++)
69 d[i][j] = d[j][i] = DIS(i, j);
70 }
71 scanf("%d", &m);
72 for (int i = 0; i < m; i++)
73 {
74 int v1, v2;
75 scanf("%d%d", &v1, &v2);
76 d[v1][v2] = d[v2][v1] = 0;
77 }
78 prim();
79 return 0;
80 }
POJ2349(ZOJ1914)-Arctic Network
1 //Prim
2 //POJ2349-ZOJ1914
3 //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
4 //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)
5 //Time:47Ms Memory:2164K
6 #include<iostream>
7 #include<cstring>
8 #include<cstdio>
9 #include<cmath>
10 #include<algorithm>
11 using namespace std;
12
13 #define MAX 501
14 #define INF 0x3f3f3f3f
15 #define POW2(x) ((x)*(x))
16 #define DIS(i,j) (sqrt(POW2(p[i][0] - p[j][0]) + POW2(p[i][1] - p[j][1])))
17
18 int n, m;
19 double p[MAX][2]; //point
20 double d[MAX][MAX]; //distance
21 double lowcost[MAX];
22 bool v[MAX];
23
24 void prim()
25 {
26 memset(lowcost, 0, sizeof(lowcost));
27 memset(v, false, sizeof(v));
28 v[0] = true;
29 for (int i = 1; i < m; i++)
30 lowcost[i] = d[i][0];
31 for (int i = 1; i < m; i++)
32 {
33 int mind = INF;
34 int k;
35 for (int j = 1; j < m; j++)
36 {
37 if (!v[j] && mind > lowcost[j])
38 {
39 mind = lowcost[j];
40 k = j;
41 }
42 }
43 v[k] = true;
44 for (int j = 1; j < m; j++)
45 if(!v[j]) lowcost[j] = min(d[k][j], lowcost[j]);
46 }
47 }
48
49 int main()
50 {
51 int T;
52 scanf("%d", &T);
53 while (T--)
54 {
55 scanf("%d%d", &n, &m);
56 for (int i = 0; i < m; i++)
57 {
58 scanf("%lf%lf", &p[i][0], &p[i][1]);
59 for (int j = 0; j < i; j++)
60 d[i][j] = d[j][i] = DIS(i, j);
61 }
62
63 prim();
64 sort(lowcost, lowcost + m);
65 printf("%.2lf\n", lowcost[m - n]);
66 //G++需要使用printf("%.2f\n", lowcost[m-n]);
67 //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
68 }
69 return 0;
70 }
POJ3026-Borg Maze
1 //Prim+BFS
2 //总是心想着要创造一个新算法，结果越想越麻烦...
3 //保险做法:找出每个点间的距离，再进行Prim
4 //Time:79Ms Memory:292K
5 #include<iostream>
6 #include<cstring>
7 #include<cstdio>
8 #include<queue>
9 #include<algorithm>
10 using namespace std;
11
12 #define MAX 125 //MAX 50的话会RE或WA(博主在55RE，在100WA)
13
14 struct Point{
15 int x, y;
16 int step;
17 }p;
18
19 int r, c;
20 char board[MAX][MAX];
21 int d[MAX][MAX];
22 int num[MAX][MAX], cnt;
23 int lowcost[MAX];
24 bool v[MAX][MAX];
25 int mov[4][2] = { {1,0}, {-1,0}, {0,1}, {0,-1} };
26
27 void bfs(Point p)
28 {
29 memset(v, false, sizeof(v));
30 v[p.x][p.y] = true;
31 int np = num[p.x][p.y];
32 queue<Point> q;
33 p.step = 0;
34 q.push(p);
35 while (!q.empty())
36 {
37 Point cur = q.front();
38 q.pop();
39 for (int i = 0; i < 4; i++)
40 {
41 Point t = cur;
42 t.x += mov[i][0];
43 t.y += mov[i][1];
44 if (t.x > 0 && t.y > 0 && t.x < r && t.y < c && !v[t.x][t.y])
45 {
46 if (board[t.x][t.y] == ‘#‘) continue;
47 int nt = num[t.x][t.y];
48 t.step++;
49 v[t.x][t.y] = true;
50 if (board[t.x][t.y] == ‘A‘ || board[t.x][t.y] == ‘S‘)
51 d[nt][np] = t.step;
52 q.push(t);
53 }
54 }
55 }
56 }
57
58 void prim()
59 {
60 int v[MAX];
61 memset(v, false, sizeof(v));
62 memset(lowcost, 0x3f, sizeof(lowcost));
63 v[0] = true;
64 for (int i = 1; i < cnt; i++)
65 lowcost[i] = d[i][0];
66
67 int minv = 0;
68 for (int i = 1; i < cnt; i++)
69 {
70 int mind = 0x3f3f3f3f;
71 int k;
72 for (int j = 1; j < cnt; j++)
73 {
74 if (!v[j] && mind > lowcost[j])
75 {
76 mind = lowcost[j];
77 k = j;
78 }
79 }
80 minv += mind;
81 v[k] = true;
82 for (int j = 1; j < cnt; j++)
83 if (!v[j]) lowcost[j] = min(lowcost[j], d[k][j]);
84 }
85 printf("%d\n", minv);
86 }
87
88 int main()
89 {
90 int T;
91 scanf("%d", &T);
92 while (T--)
93 {
94 cnt = 0;
95 scanf("%d%d", &c, &r);
96 gets_s(board[0]);
97 for (int i = 0; i < r; i++)
98 {
99 gets_s(board[i], MAX);
100 for (int j = 0; j < c; j++)
101 if (board[i][j] == ‘S‘ || board[i][j] == ‘A‘)
102 num[i][j] = cnt++;
103 }
104 for (int i = 0; i < r; i++)
105 for (int j = 0; j < c;j++)
106 if (board[i][j] == ‘S‘ || board[i][j] == ‘A‘)
107 {
108 p.x = i; p.y = j;
109 bfs(p);
110 }
111 prim();
112 }
113 return 0;
114 }
时间： 2024-10-02 19:51:58