BZOJ4031: [HEOI2015]小Z的房间

n,m<=9,n*m的网格图,相邻的.可连边,问把所有的.连成一棵树有多少方案,%1e9。

直接矩阵树,然而高斯消元时模数不是质数没法直接除,所以利用行列式的性质,某一行乘某个数加到另一行上,这样辗转相除。

 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<algorithm>
 4 #include<stdlib.h>
 5 //#include<queue>
 6 #include<math.h>
 7 //#include<time.h>
 8 //#include<iostream>
 9 using namespace std;
10 
11 int n,m,tot=0;
12 char mp[13][13];
13 int mat[111][111],who[111][111];
14 const int mod=1e9;
15 
16 int gauss(int n)
17 {
18     for (int i=1;i<=n;i++)
19         for (int j=1;j<=n;j++)
20             mat[i][j]+=mat[i][j]<0?mod:0;
21     int ans=1;
22     for (int i=1;i<=n;i++)
23         for (int j=i+1;j<=n;j++)
24         {
25             int A=mat[i][i],B=mat[j][i];
26             while (B)
27             {
28                 int t=A/B; A%=B; swap(A,B);
29                 for (int k=i;k<=n;k++)
30                     mat[i][k]-=1ll*mat[j][k]*t%mod,
31                     mat[i][k]+=mat[i][k]<0?mod:0;
32                 for (int k=i;k<=n;k++) {int t=mat[j][k]; mat[j][k]=mat[i][k]; mat[i][k]=t;}
33                 ans=ans==1?mod-1:1;
34             }
35         }
36     for (int i=1;i<=n;i++) ans=1ll*ans*mat[i][i]%mod;
37     return ans;
38 }
39 
40 int main()
41 {
42     scanf("%d%d",&n,&m);
43     for (int i=1;i<=n;i++) scanf("%s",mp[i]+1);
44     for (int i=1;i<=n;i++)
45         for (int j=1;j<=m;j++)
46             if (mp[i][j]=='.') who[i][j]=++tot;
47     for (int i=1;i<=n;i++)
48         for (int j=1;j<=m;j++)
49             if (mp[i][j]=='.')
50             {
51                 if (mp[i+1][j]=='.') mat[who[i][j]][who[i+1][j]]--,mat[who[i][j]][who[i][j]]++;
52                 if (mp[i][j+1]=='.') mat[who[i][j]][who[i][j+1]]--,mat[who[i][j]][who[i][j]]++;
53                 if (mp[i-1][j]=='.') mat[who[i][j]][who[i-1][j]]--,mat[who[i][j]][who[i][j]]++;
54                 if (mp[i][j-1]=='.') mat[who[i][j]][who[i][j-1]]--,mat[who[i][j]][who[i][j]]++;
55             }
56     printf("%d\n",gauss(tot-1));
57     return 0;
58 }
View Code

 

posted @ 2017-12-28 09:06  Blue233333  阅读(164)  评论(0编辑  收藏  举报