UVA10766-Organising the Organisation（生成树计数+Matrix-tree定理）

来源：互联网发布：二分查找算法java 编辑：程序博客网时间：2024/06/16 13:02

题目链接

http://acm.hust.edu.cn/vjudge/contest/67265#problem/I

题意

给定一张图，并且确定根节点，求生成树的个数

思路

因为是无向图，所以哪个节点作为根节点都没有影响，因此题目转化为无向图的生成树计数
Matrix-tree定理：
对于图G，设D[G]为G的度数矩阵，A[G]为G的邻接矩阵，则C[G] = D[G] - A[G]；并且生成树的个数 = C[G]的任意一个n - 1阶主子式的行列式
因此只需要求出行列式即可
求行列式采用消元法（还没有弄清楚为什么用long double会WA）

代码

#include <iostream>#include <cstring>#include <stack>#include <vector>#include <set>#include <map>#include <cmath>#include <queue>#include <sstream>#include <iomanip>#include <fstream>#include <cstdio>#include <cstdlib>#include <climits>#include <deque>#include <bitset>#include <algorithm>using namespace std;#define PI acos(-1.0)#define LL long long#define PII pair<int, int>#define PLL pair<LL, LL>#define mp make_pair#define IN freopen("in.txt", "r", stdin)#define OUT freopen("out.txt", "wb", stdout)#define scan(x) scanf("%d", &x)#define scan2(x, y) scanf("%d%d", &x, &y)#define scan3(x, y, z) scanf("%d%d%d", &x, &y, &z)#define sqr(x) (x) * (x)#define pr(x) cout << #x << " = " << x << endl#define lc o << 1#define rc o << 1 | 1#define pl() cout << endl#define CLR(a, x) memset(a, x, sizeof(a))#define FILL(a, n, x) for (int i = 0; i < n; i++) a[i] = xconst int maxn = 55;int n, m, k;LL C[maxn][maxn], A[maxn][maxn];LL det(LL a[][maxn], int n) {    int i, j, k, r;    LL res = 1;    for (i = 0; i < n; i++) {        for (j = i + 1; j < n; j++) {            while (a[j][i]) {                LL f = a[i][i] / a[j][i];                for (int k = i; k < n; k++) a[i][k] -= f * a[j][k];                for (int k = i; k < n; k++) swap(a[i][k], a[j][k]);                res = -res;            }        }        if (a[i][i] == 0) return 0;        res *= a[i][i];    }    return res < 0 ? -res : res;}void init() {    memset(C, 0, sizeof(C));    for (int i = 0; i < n; i++) {        for (int j = 0; j < n; j++) {            A[i][j] = 1;        }    }}int main() {    while (~scan3(n, m, k)) {        init();        for (int i = 0; i < m; i++) {            int x, y;            scan2(x, y);            x--; y--;            A[x][y] = A[y][x] = 0;        }        for (int i = 0; i < n; i++) {            for (int j = 0; j < n; j++) {                if (i != j && A[i][j]) {                    C[i][i]++;                    C[i][j] = -1;                }            }        }        LL ans = det(C, n - 1);        printf("%lld\n", ans);    }    return 0;}

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