http://acm.timus.ru/problem.aspx?space=1&num=1627

　　生成樹計(jì)數(shù)的題，直接用Matrix-Tree定理就可以解決問題了。

代碼如下：

View Code 1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 5 using namespace std; 6 7 const int maxSize = 100; 8 const int initMod = 1e9; 9 typedef long long ll; 10 int curSize = maxSize; 11 12 struct Mat { 13 ll val[maxSize][maxSize]; 14 15 Mat (ll one = 0) { 16 for (int i = 0; i < curSize; i++) { 17 for (int j = 0; j < curSize; j++) { 18 val[i][j] = 0; 19 } 20 val[i][i] = one; 21 } 22 } 23 24 void print() { 25 for (int i = 0; i < curSize; i++) { 26 for (int j = 0; j < curSize; j++) { 27 printf("%lld ", val[i][j]); 28 } 29 puts(""); 30 } 31 puts("~~~"); 32 } 33 } mat; 34 35 const int maxn = 12; 36 char mp[maxn][maxn]; 37 38 void convert(int l, int r) { 39 int mark = 0; 40 41 for (int i = 0; i < l; i++) { 42 for (int j = 0; j < r; j++) { 43 mp[i][j] = (mp[i][j] == '.') ? ++mark : 0; 44 } 45 } 46 // printf("mark %d\n", mark); 47 48 curSize = mark + 1; 49 mat = Mat(); 50 51 for (int i = 0; i < l; i++) { 52 for (int j = 0; j < r; j++) { 53 if (!mp[i][j]) continue; 54 if (i && mp[i - 1][j]) { 55 mat.val[mp[i][j]][mp[i - 1][j]]--; 56 mat.val[mp[i][j]][mp[i][j]]++; 57 } 58 if (j && mp[i][j - 1]) { 59 mat.val[mp[i][j]][mp[i][j - 1]]--; 60 mat.val[mp[i][j]][mp[i][j]]++; 61 } 62 if (i != l - 1 && mp[i + 1][j]) { 63 mat.val[mp[i][j]][mp[i + 1][j]]--; 64 mat.val[mp[i][j]][mp[i][j]]++; 65 } 66 if (j != r - 1 && mp[i][j + 1]) { 67 mat.val[mp[i][j]][mp[i][j + 1]]--; 68 mat.val[mp[i][j]][mp[i][j]]++; 69 } 70 } 71 } 72 // puts("~~~"); 73 // mat.print(); 74 } 75 76 ll gcd(ll a, ll b) { 77 return b ? gcd(b, a % b) : a; 78 } 79 80 ll lcm(ll a, ll b) { 81 return a / gcd(a, b) * b; 82 } 83 84 ll cal() { 85 ll ret = 1ll; 86 87 curSize--; 88 for (int p = 1, i, j; p < curSize; p++) { 89 for (i = p + 1; i < curSize; i++) { 90 while (mat.val[i][p]) { 91 int tmp = mat.val[p][p] / mat.val[i][p]; 92 93 for (j = p; j < curSize; j++) { 94 mat.val[p][j] = ((mat.val[p][j] - mat.val[i][j] * tmp) % initMod + initMod) % initMod; 95 swap(mat.val[i][j], mat.val[p][j]); 96 } 97 ret = -ret; 98 } 99 } 100 ret *= mat.val[p][p]; 101 ret = (ret % initMod + initMod) % initMod; 102 // mat.print(); 103 } 104 105 return ret; 106 } 107 108 int main() { 109 // freopen("in", "r", stdin); 110 // freopen("out", "w", stdout); 111 int n, m; 112 113 while (~scanf("%d%d", &n, &m)) { 114 for (int i = 0; i < n; i++) { 115 scanf("%s", mp[i]); 116 } 117 convert(n, m); 118 printf("%lld\n", cal()); 119 } 120 121 return 0; 122 }

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——written by Lyon

轉(zhuǎn)載于:https://www.cnblogs.com/LyonLys/archive/2012/10/29/Ural_1627_Lyon.html

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