标签：aced 存在 namespace edr ber ext any least hat
Join
题目链接：https://vjudge.net/problem/URAL-1627
Description:
Businessman Petya recently bought a new house. This house has one floor with n × m square rooms, placed in rectangular lattice. Some rooms are pantries and the other ones are bedrooms. Now he wants to join all bedrooms with doors in such a way that there will be exactly one way between any pair of them. He can make doors only between neighbouring bedrooms (i.e. bedrooms having a common wall). Now he wants to count the number of different ways he can do it.
Input:
First line contains two integers n and m (1 ≤ n, m ≤ 9) — the number of lines and columns in the lattice. Next n lines contain exactly m characters representing house map, where "." means bedroom and "*" means pantry. It is guaranteed that there is at least one bedroom in the house.
Output:
Output the number of ways to join bedrooms modulo 10 9.
Sample Input:
2 2 .* *.
Sample Output:
0
题意：
给出一个n*m的矩阵，然后"*"表示障碍物。现在可以在两个"."之间搭桥，问有多少种方式将所有的"."连通。
题解：
直接给所有能够搭桥的点建边，然后就是个生成树计数问题了。
代码如下：
#include <cstdio> #include <cstring> #include <algorithm> #include <iostream> #include <cmath> using namespace std; typedef long long ll; const int N = 105,MOD = 1e9; char mp[N][N]; int g[N][N],num[N][N]; ll b[N][N]; int tot=0; ll Det(int n){ int i,j,k; ll ret = 1; if(n==0) return 0; for(i=2;i<=n;i++){ for(j = i+1;j <= n;j++){ while(b[j][i]){ ll tmp=b[i][i]/b[j][i];//不存在除不尽的情况 for(k = i;k <= n;k++) b[i][k] = ((b[i][k] - tmp*b[j][k])%MOD+MOD)%MOD; for(k=i;k<=n;k++) swap(b[i][k],b[j][k]); ret = -ret; } } if(!b[i][i]) return 0; ret = ret * b[i][i]%MOD; if(ret<0) ret+=MOD; } if(ret < 0) ret += MOD; return ret; } int main(){ int n,m; scanf("%d%d",&n,&m); for(int i=1;i<=n;i++){ scanf("%s",mp[i]+1); for(int j=1;j<=m;j++){ if(mp[i][j]==‘.‘) num[i][j]=++tot; } } for(int i=1;i<=n;i++){ for(int j=1;j<=m;j++){ if(mp[i][j]==‘*‘) continue ; if(j>1&&mp[i][j-1]==‘.‘) g[num[i][j-1]][num[i][j]]=1; if(i<n&&mp[i+1][j]==‘.‘) g[num[i+1][j]][num[i][j]]=1; } } for(int i=1;i<=tot;i++){ for(int j=1;j<=tot;j++){ if(g[i][j]){ b[i][i]++;b[j][j]++; b[i][j]=b[j][i]=-1; } } } cout<<Det(tot); return 0; }
标签：aced 存在 namespace edr ber ext any least hat
原文地址：https://www.cnblogs.com/heyuhhh/p/10392763.html
