Matrix_tree Theorem 矩阵树定理学习笔记

Matrix_tree Theorem:

给定一个无向图, 定义矩阵A

A[i][j] = - (<i, j>之间的边数)

A[i][i] = 点i的度数

其生成树的个数等于 A的任意n - 1阶主子式的值。

关于定理的相关证明可以看这篇文章, 讲得非常详细, 耐心看就能看懂：

关于求行列式, 可以用高斯消元。如果是模域下求行列式, 可以用欧几里得算法。具体实现看这篇文章

模域下求行列式模板题：SPOJ DETER3

代码：

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <queue>
 4 #include <algorithm>
 5 #include <cstring>
 6 #include <set>
 7 #include <cmath>
 8 using namespace std;
 9 
10 #define N 220
11 #define M 400010
12 typedef long long ll;
13 
14 const int Mod=1000000007;
15 const double eps = 1e-9;
16 
17 ll Solve(ll a[N][N], int n, ll mod)
18 {
19     ll res = 1;
20     for (int i = 1; i <= n; ++i)
21     {
22         for (int j = i; j <= n; ++j)
23         {
24             if (a[j][i] < 0)
25             {
26                 res *= -1;
27                 for (int k = i; k <= n; ++k)
28                     a[j][k] *= -1;
29             }
30         }
31         int j;
32         for (j = i; j <= n && !a[j][i]; ++j);
33         if (j > n) return 0;
34         
35         if (j != i)    
36         {
37             res = -res;
38             for (int k = i; k <= n; ++k) swap(a[i][k], a[j][k]);
39         }
40         
41         for (j = i + 1; j <= n; ++j)
42         {
43             while (a[j][i])
44             {
45                 ll d = a[i][i] / a[j][i];
46                 for (int k = i; k <= n; ++k) a[i][k] -= d * a[j][k] % mod, a[i][k] %= mod;
47                 for (int k = i; k <= n; ++k) swap(a[i][k], a[j][k]);
48                 res = -res;
49             }
50         }
51         res = res * a[i][i] % mod;
52     }
53     if (res < 0) res += mod;
54     return res;
55 }
56 
57 int main()
58 {
59     //freopen("in.in","r",stdin);
60     //freopen("out.out","w",stdout);
61     
62     int n; ll mod; 
63     while (scanf("%d %lld", &n, &mod) != EOF)
64     {
65         ll a[N][N];
66         for (int i = 1; i <= n; ++i)
67             for (int j = 1; j <= n; ++j)
68                 scanf("%lld", &a[i][j]);
69         printf("%lld\n", Solve(a, n, mod));
70     }
71     
72     return 0;
73 }

View Code

下面给出一些应用(练习题):

应用一：SPOJ HIGH

模板题：给出一个无向图, 求生成树个数。

代码：

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <queue>
 4 #include <algorithm>
 5 #include <cstring>
 6 #include <set>
 7 #include <cmath>
 8 using namespace std;
 9 
10 #define N 13
11 #define M 400010
12 typedef long long ll;
13 
14 const int Mod=1000000007;
15 const double eps = 1e-9;
16 
17 double Solve(int n, double a[N][N])
18 {
19     if (n == 0) return 1;
20     
21     /*for (int i = 1; i <= n; ++i)
22         for (int j = 1; j <= n; ++j)
23             printf("%.0lf%c", a[i][j], j == n? '\n':' ');*/
24     
25     double res = 1;
26     for (int i = 1; i <= n; ++i)
27     {
28         int j;
29         for (j = i; j <= n && fabs(a[j][i]) < eps; ++j);
30         if (j > n) return 0;
31         if (j != i) for (int k = i; k <= n; ++k) swap(a[i][k], a[j][k]);
32         
33         for (int j = i + 1; j <= n; ++j)
34         {
35             double f = a[j][i] / a[i][i];
36             for (int k = i; k <= n; ++k)
37                 a[j][k] -= f * a[i][k];
38         }
39         res *= a[i][i];
40     }
41 
42     return res;
43 }
44 
45 int main()
46 {
47     //freopen("in.in","r",stdin);
48     //freopen("out.out","w",stdout);
49     
50     int T, n, m, x, y;
51     scanf("%d", &T);
52     while (T--)
53     {
54         double a[N][N] = {0};
55         scanf("%d %d", &n, &m);
56         for (int i = 1; i <= m; ++i)
57         {
58             scanf("%d %d", &x, &y);
59             a[x][y]--, a[y][x]--;
60             a[x][x]++, a[y][y]++;
61         }
62         printf("%.0lf\n", Solve(n - 1, a));
63     }
64     
65     return 0;
66 }

View Code

应用二：BZOJ 4031

构图后求生成树个数 mod 一个数。

  1 #include <cstdio>
  2 #include <iostream>
  3 #include <queue>
  4 #include <algorithm>
  5 #include <cstring>
  6 #include <set>
  7 #include <cmath>
  8 using namespace std;
  9  
 10 #define N 120
 11 #define M 400010
 12 typedef long long ll;
 13  
 14 const int Mod=1000000007;
 15 const double eps = 1e-9;
 16  
 17 ll Solve(ll a[N][N], int n, ll mod)
 18 {
 19     if (n == 0) return 1;
 20     ll res = 1;
 21     for (int i = 1; i <= n; ++i)
 22     {
 23         //for (int p = 1; p <= n; ++p)
 24         //  for (int q = 1; q <= n; ++q)
 25         //      printf("%lld%c", a[p][q], q == n? '\n':' ');
 26          
 27         for (int j = i; j <= n; ++j) 
 28         {
 29             if (a[j][i] < 0)
 30             {
 31                 res = -res;
 32                 for (int k = i; k <= n; ++k) a[j][k] *= -1;
 33             }
 34         }   
 35          
 36         int j;
 37         for (j = i; j <= n && !a[j][i]; ++j);
 38         if (j > n) return 0;
 39          
 40         if (j != i)
 41         {
 42             res = -res;
 43             for (int k = i; k <= n; ++k) swap(a[i][k], a[j][k]);
 44         }
 45          
 46          
 47         for (j = i + 1; j <= n; ++j)
 48         {
 49             while (a[j][i])
 50             {
 51                 ll d = a[i][i] / a[j][i];
 52                 for (int k = i; k <= n; ++k) a[i][k] -= d * a[j][k] % mod, a[i][k] %= mod;
 53                 res = -res;
 54                 for (int k = i; k <= n; ++k) swap(a[i][k], a[j][k]);
 55             }
 56         }
 57         res = res * a[i][i] % mod;
 58     //  printf("res = %lld\n", res);
 59     }
 60     if (res < 0) res += mod;
 61 //  cout << "aa= "<<res <<endl;
 62     return res;
 63 }
 64  
 65 int main()
 66 {
 67     //freopen("in.in","r",stdin);
 68     //freopen("out.out","w",stdout);
 69      
 70     int n, m; ll mod = 1000000000; 
 71     char mp[11][11]; int tot = 0;
 72     int id[11][11];
 73     scanf("%d %d", &n, &m);
 74     for (int i = 1; i <= n; ++i)
 75     {
 76         for (int j = 1; j <= m; ++j)
 77         {
 78             scanf(" %c", &mp[i][j]);
 79             if (mp[i][j] == '.') id[i][j] = ++tot;
 80         }
 81     }
 82     ll a[N][N] = {0};
 83     for (int i = 1; i < n; ++i)
 84     {
 85         for (int j = 1; j <= m; ++j)
 86         {
 87             if (mp[i][j] == '.' && mp[i + 1][j] == '.')
 88             {
 89                 int x = id[i][j], y = id[i + 1][j];
 90                 a[x][y]--, a[y][x]--;
 91                 a[x][x]++, a[y][y]++;
 92             }
 93         }
 94     }
 95     for (int i = 1; i <= n; ++i)
 96     {
 97         for (int j = 1; j < m; ++j)
 98         {
 99             if (mp[i][j] == '.' && mp[i][j + 1] == '.')
100             {
101                 int x = id[i][j], y = id[i][j + 1];
102                 a[x][y]--, a[y][x]--;
103                 a[x][x]++, a[y][y]++;
104             }
105         }
106     }
107     printf("%lld\n", Solve(a, tot - 1, mod));
108     return 0;
109 }

View Code

应用三： BZOJ 2467

这题数据范围比较小,可以暴力建图然后跑Matrix tree。

另外可以直接推公式：

一共有4n个点, 5n条边, 所以要删去n - 1条边, 然后可以发现每个五边形外面的4条边最多只能删一条。

根据鸽笼原理合法的解一定是有一个五边形删去了里面的那条边和外面的某条边, 其余的五边形删去了任意一条边。

所以答案就是$4*n*5^{n-1}$

Matrix tree 代码：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 #include <vector>
 6 #include <cmath>
 7 #include <queue>
 8 #include <map>
 9 using namespace std;
10  
11 #define X first
12 #define Y second
13 #define N 420
14 #define M 11
15  
16 typedef long long ll;
17 const int Mod = 1000000007;
18 const int INF = 1 << 30;
19  
20 void Add(int x, int y, int &tot, int a[N][N])
21 {
22     a[x][tot + 1] = a[tot + 1][x] = -1;
23     a[tot + 1][tot + 2] = a[tot + 2][tot + 1] = -1;
24     a[tot + 2][tot + 3] = a[tot + 3][tot + 2] = -1;
25     a[tot + 3][y] = a[y][tot + 3] = -1;
26     a[tot + 1][tot + 1] = a[tot + 2][tot + 2] = a[tot + 3][tot + 3] = 2;
27     a[x][x] = a[y][y] = 4; tot += 3;
28     a[x][y]--,a[y][x]--;
29 }
30  
31 int Solve(int n, int a[N][N], int mod)
32 {
33  
34     if (n == 0) return 1;
35     int res = 1;
36     for (int i = 1; i <= n; ++i)
37     {
38         for (int j = i; j <= n; ++j)
39         {
40             if (a[j][i] < 0)
41             {
42                 res = -res;
43                 for (int k = i; k <= n; ++k) a[j][k] = -a[j][k];
44             }
45         }
46         //cout << i << endl;
47         //for (int p = 1; p <= n; ++p)
48         //  for (int q = 1; q <= n; ++q)
49         //  printf("%d%c", a[p][q], q == n? '\n':' ');
50         //printf("\n");
51         int j;
52         for (j = i; j <= n && !a[j][i]; ++j);
53         if (j > n) return 0;
54         if (i != j)
55         {
56             res = -res;
57             for (int k = i; k <= n; ++k) swap(a[i][k], a[j][k]);
58         }
59          
60         for (j = i + 1; j <= n; ++j)
61         {
62             while (a[j][i])
63             {
64                 int d = a[i][i] / a[j][i];
65                 for (int k = i; k <= n; ++k) a[i][k] -= d * a[j][k] % mod, a[i][k] %= mod;
66                 res = -res;
67                 for (int k = i; k <= n; ++k) swap(a[i][k], a[j][k]);
68             }
69         }
70         res = res * a[i][i] % mod;
71     }
72     if (res < 0) res += mod;
73     return res;
74 }
75  
76 int main()
77 {
78     //freopen("in.in","r",stdin);
79     //freopen("out.out","w",stdout);
80      
81     int T; scanf("%d", &T);
82     while (T--)
83     {
84         int n, mod = 2007, tot, a[N][N] = {0}; scanf("%d", &n); tot = n;
85         for (int i = 1; i < n; ++i) Add(i, i + 1, tot, a);
86         Add(n, 1, tot, a); 
87         printf("%d\n", Solve(tot - 1, a, mod));
88     }
89     return 0;
90 }

View Code

应用四：BZOJ 1016

题目大意：求最小生成树个数。

性质一：无向图所有MST中，相同权值的边数一样多。

证明看https://blog.sengxian.com/solutions/bzoj-1016

性质二：对于任意MST，加入所有权值<=w的边后, 形成的森林连通性相同。

证明：

考虑Kruskal算法的过程，我们首先会尽可能多的加入权值最小的边。这个过程相当于拿出所有权值最小的边，然后任意求一颗生成树，因此我们可以知道，能加入的权值最小的边的数量是一定的，而且加入这些边之后形成的森林连通性相同。

结合性质一，对于任意一棵MST，因为它包含的权值最小的边数和做Kruskal算法求出的MST包含的边数是一样的，这些边又不能形成环，因此这些边形成的森林和做Kruskal时形成的森林连通性是一样的。对于任意MST，加入所有权值最小的边后, 形成的森林连通性相同。

然后我们考虑把已经形成的联通块缩点，考虑所有权值第二小的边，重复上面的过程，可以证明对于任意MST，加入所有权值<=w的边后, 形成的森林连通性相同。

所以我们的算法就可以模拟这个过程, 把边按权值从小到大排好序, 每次加入权值相同的所有边, 形成一些连通块，然后对于每个连通块, 跑Matrix tree 求出形成这个连通块有多少种方案。统计好之后每个连通块缩点, 进行下一种权值的边的加边操作。代码实现起来还是挺多细节的。

代码：

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <algorithm>
  5 #include <vector>
  6 #include <cmath>
  7 #include <queue>
  8 #include <map>
  9 using namespace std;
 10 
 11 #define X first
 12 #define Y second
 13 #define N 120
 14 #define M 1010
 15 
 16 typedef long long ll;
 17 const int Mod = 1000000007;
 18 const int INF = 1 << 30;
 19 
 20 struct Edge
 21 {
 22     int x, y, z;
 23     bool operator < (const Edge &t)const{return z < t.z;}
 24 }e[M];
 25 
 26 
 27 int Solve(int n, int a[N][N], int mod)
 28 {
 29     if (n == 0) return 1;
 30     int res = 1;
 31     for (int i = 0; i < n; ++i)
 32     {
 33         for (int j = i; j < n; ++j)
 34         {
 35             if (a[j][i] < 0)
 36             {
 37                 res = -res;
 38                 for (int k = i; k < n; ++k) a[j][k] = -a[j][k];
 39             }
 40         }
 41         int j;
 42         for (j = i; j < n && !a[j][i]; ++j);
 43         if (j == n) return 0;
 44         if (i != j)
 45         {
 46             res = -res;
 47             for (int k = i; k < n; ++k) swap(a[i][k], a[j][k]);
 48         }
 49         
 50         for (j = i + 1; j < n; ++j)
 51         {
 52             while (a[j][i])
 53             {
 54                 int d = a[i][i] / a[j][i];
 55                 for (int k = i; k < n; ++k) a[i][k] -= d * a[j][k] % mod, a[i][k] %= mod;
 56                 res = -res;
 57                 for (int k = i; k < n; ++k) swap(a[i][k], a[j][k]);
 58             }
 59         }
 60         res = res * a[i][i] % mod;
 61     }
 62     if (res < 0) res += mod;
 63     return res;
 64 }
 65 
 66 
 67 int father[N], id[N], pa[N], num[N];
 68 vector<int> lis[N];
 69 vector<pair<int, int> > ed[N];
 70 
 71 int Find(int x)
 72 {
 73     if (father[x] == x) return x;
 74     father[x] = Find(father[x]);
 75     return father[x];
 76 }
 77 
 78 void Merge(int x, int y)
 79 {
 80     x = Find(x), y = Find(y);
 81     if (x != y) father[x] = y;
 82 }
 83 
 84 int main()
 85 {
 86     //freopen("in.in","r",stdin);
 87     //freopen("out.out","w",stdout);
 88     
 89     int n, m, mod = 31011; 
 90     scanf("%d %d", &n, &m);
 91     for (int i = 1; i <= m; ++i) scanf("%d %d %d", &e[i].x, &e[i].y, &e[i].z);
 92     sort(e + 1, e + m + 1);
 93     
 94     int res = 1, block = n;
 95     for (int i = 1; i <= n; ++i) id[i] = i;
 96     for (int l = 1, r; l <= m;)
 97     {
 98         for (r = l; r < m && e[r + 1].z == e[l].z; ++r);
 99         for (int i = 1; i <= block; ++i) father[i] = i;
100         for (r = l; r < m && e[r + 1].z == e[l].z; ++r);
101         for (int i = l; i <= r; ++i) Merge(id[e[i].x], id[e[i].y]);
102         
103         int tot = 0;
104         for (int i = 1; i <= block; ++i) if (father[i] == i) pa[i] = ++tot;
105         for (int i = 1; i <= block; ++i) pa[i] = pa[Find(i)];
106         for (int i = 1; i <= block; ++i) lis[pa[i]].push_back(i), num[i] = lis[pa[i]].size() - 1;
107         for (int i = l; i <= r; ++i)
108         {
109             int x = id[e[i].x], y = id[e[i].y];
110             if (x == y) continue;
111             ed[pa[x]].push_back(make_pair(num[x], num[y]));
112         }
113         for (int i = 1; i <= tot; ++i)
114         {
115             int a[N][N] = {0}, x, y;
116             for (int j = 0; j < ed[i].size(); ++j)
117             {
118                 x = ed[i][j].X, y = ed[i][j].Y;
119                 a[x][x]++, a[y][y]++;
120                 a[x][y]--, a[y][x]--;
121             }
122             res = res * Solve(lis[i].size() - 1, a, mod) % mod;
123         }
124         
125         for (int i = 1; i <= n; ++i) id[i] = pa[id[i]];
126         for (int i = 1; i <= tot; ++i) lis[i].clear(), ed[i].clear();
127         block = tot;  l = r + 1;
128     }
129     if (block > 1) puts("0");
130     else printf("%d\n", res);
131     return 0;
132 }

View Code

应用五： BZOJ 3534 Matrix Tree Theorem 的扩展, 非常精彩的题。

题目大意：

给出一个无向图, 两点之间的连边会有一个概率，求连成一颗树的概率。

这个题如果没有看过Matrix Tree Theorem定理的证明,只是记住结论应该是做不出来的。。。

先来看一个简化版本：

给出一个无向图, 定义它的一棵生成树的权值为所有边权的乘积。求所有生成树的权值和。 ( 原题还要考虑一些边不选的概率这题只靠考虑选的边）

参考最前面给出的证明Matrix Tree Theorem定理的文章。

原来是

A[i][j] = - (<i, j>之间的边数)

A[i][i] = 点i的度数

现在改成

A[i][j] = - (<i, j>之间的所有边权和)

A[i][i] = 和i相连的所有边的边权和

修改关联矩阵B的定义, 把1 改成 $e_j$的边权开根号后的值。

这里也做相应的修改, 把 -1 改成 -<i,j>之间的边权和, the degree of i 改成和i相连的所有边的边权和。

做了以上修改之后刚好就是所选的生成树的边权的乘积。

所以用修改后的A数组跑Matrix Tree就可以解决这个问题了。

当然原题还要乘上其他边不选的概率。再对A数组做点小修改就好了。具体实现看代码：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 #include <vector>
 6 #include <cmath>
 7 #include <queue>
 8 #include <map>
 9 using namespace std;
10  
11 #define X first
12 #define Y second
13 #define N 100
14 #define M 11
15  
16 typedef long long ll;
17 const int Mod=1000000007;
18 const int INF=1<<30;
19  
20 const double eps = 1e-10;
21  
22 double Solve(double a[N][N], int n)
23 {
24     if (n == 0) return 1;
25     double res = 1;
26     for (int i = 1; i <= n; ++i)
27     {
28         int j = i;
29         for (int k = i + 1; k <= n ; ++k) if (fabs(a[k][i]) > fabs(a[j][i])) j = k;
30         if (fabs(a[j][i]) < eps) return 0;
31         if (i != j)
32         {
33             res = -res;
34             for (int k = i; k <= n; ++k) swap(a[i][k], a[j][k]);
35         }
36          
37         for (j = i + 1; j <= n; ++j)
38         {
39             double f = a[j][i] / a[i][i];
40             for (int k = i; k <= n; ++k) a[j][k] -= f * a[i][k];
41         }
42         res *= a[i][i];
43     }
44     return res;
45 }
46  
47 int main()
48 {
49     //freopen("in.in","r",stdin);
50     //freopen("out.out","w",stdout);
51      
52     int n; 
53     double a[N][N] = {0}, res = 1;
54      
55     scanf("%d", &n);
56     for (int i = 1; i <= n; ++i)
57     {
58         for (int j = 1; j <= n; ++j)
59         {
60             scanf("%lf", &a[i][j]);
61             if (i == j) continue;
62             if (i < j && fabs(a[i][j] - 1) > eps) res *= 1 - a[i][j]; 
63             if (fabs(a[i][j] - 1) > eps) a[i][j] = -a[i][j] / (1 - a[i][j]); 
64             else a[i][j] = -a[i][j];
65         }
66     }
67     for (int i = 1; i <= n; ++i)
68         for (int j = 1; j <= n; ++j)
69             if (i != j) a[i][i] -= a[i][j];
70     printf("%.8lf\n", fabs(res * Solve(a, n - 1)));
71     return 0;
72 }

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posted @ 2017-05-05 11:58 lzw4896s 阅读(411) 评论(0) 编辑收藏举报

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Matrix_tree Theorem 矩阵树定理学习笔记

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