Ural 1627 Join(生成树计数)

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

　　生成树计数的题，直接用Matrix-Tree定理就可以解决问题了。

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxSize = 100;
const int initMod = 1e9;
typedef long long ll;
int curSize = maxSize;

struct Mat {
    ll val[maxSize][maxSize];

    Mat (ll one = 0) {
        for (int i = 0; i < curSize; i++) {
            for (int j = 0; j < curSize; j++) {
                val[i][j] = 0;
            }
            val[i][i] = one;
        }
    }

    void print() {
        for (int i = 0; i < curSize; i++) {
            for (int j = 0; j < curSize; j++) {
                printf("%lld ", val[i][j]);
            }
            puts("");
        }
        puts("~~~");
    }
} mat;

const int maxn = 12;
char mp[maxn][maxn];

void convert(int l, int r) {
    int mark = 0;

    for (int i = 0; i < l; i++) {
        for (int j = 0; j < r; j++) {
            mp[i][j] = (mp[i][j] == '.') ? ++mark : 0;
        }
    }
//    printf("mark %d\n", mark);

    curSize = mark + 1;
    mat = Mat();

    for (int i = 0; i < l; i++) {
        for (int j = 0; j < r; j++) {
            if (!mp[i][j]) continue;
            if (i && mp[i - 1][j]) {
                mat.val[mp[i][j]][mp[i - 1][j]]--;
                mat.val[mp[i][j]][mp[i][j]]++;
            }
            if (j && mp[i][j - 1]) {
                mat.val[mp[i][j]][mp[i][j - 1]]--;
                mat.val[mp[i][j]][mp[i][j]]++;
            }
            if (i != l - 1 && mp[i + 1][j]) {
                mat.val[mp[i][j]][mp[i + 1][j]]--;
                mat.val[mp[i][j]][mp[i][j]]++;
            }
            if (j != r - 1 && mp[i][j + 1]) {
                mat.val[mp[i][j]][mp[i][j + 1]]--;
                mat.val[mp[i][j]][mp[i][j]]++;
            }
        }
    }
//    puts("~~~");
//    mat.print();
}

ll gcd(ll a, ll b) {
    return b ? gcd(b, a % b) : a;
}

ll lcm(ll a, ll b) {
    return a / gcd(a, b) * b;
}

ll cal() {
    ll ret = 1ll;

    curSize--;
    for (int p = 1, i, j; p < curSize; p++) {
        for (i = p + 1; i < curSize; i++) {
            while (mat.val[i][p]) {
                int tmp = mat.val[p][p] / mat.val[i][p];

                for (j = p; j < curSize; j++) {
                    mat.val[p][j] = ((mat.val[p][j] - mat.val[i][j] * tmp) % initMod + initMod) % initMod;
                    swap(mat.val[i][j], mat.val[p][j]);
                }
                ret = -ret;
            }
        }
        ret *= mat.val[p][p];
        ret = (ret % initMod + initMod) % initMod;
//        mat.print();
    }

    return ret;
}

int main() {
//    freopen("in", "r", stdin);
//    freopen("out", "w", stdout);
    int n, m;

    while (~scanf("%d%d", &n, &m)) {
        for (int i = 0; i < n; i++) {
            scanf("%s", mp[i]);
        }
        convert(n, m);
        printf("%lld\n", cal());
    }

    return 0;
}

——written by Lyon