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  • BZOJ1016 && JSOI2008] 最小生成树计数

    题目链接:点击打开链接

    裸题


    #pragma comment(linker, "/STACK:1024000000,1024000000")
    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <vector>
    template <class T>
    inline bool rd(T &ret) {
    	char c; int sgn;
    	if (c = getchar(), c == EOF) return 0;
    	while (c != '-' && (c<'0' || c>'9')) c = getchar();
    	sgn = (c == '-') ?

    -1 : 1; ret = (c == '-') ? 0 : (c - '0'); while (c = getchar(), c >= '0'&&c <= '9') ret = ret * 10 + (c - '0'); ret *= sgn; return 1; } template <class T> inline void pt(T x) { if (x <0) { putchar('-'); x = -x; } if (x>9) pt(x / 10); putchar(x % 10 + '0'); } using namespace std; typedef long long ll; const int N = 105; //点的个数 const int M = 1005; //边的个数 //点标从1-n struct node { int set[N]; void init(int n) { for (int i = 0; i <= n; i++) set[i] = i; } int find(int x) { return x == set[x] ? x : set[x] = find(set[x]); } int Union(int x, int y) { int xx = find(x); int yy = find(y); if (xx == yy) return -1; set[xx] = yy; return 1; } }a, b, c; struct Node { int u, v, dis; }edge[M]; int edgenum; void add(int u, int v, int d){ Node E = { u, v, d }; edge[++edgenum] = E; } bool visit[N]; vector<int> g[N]; ll p[N][N], deg[N][N]; int cmp(Node a, Node b) { return a.dis < b.dis; } ll DET(ll a[][N], int n, ll MOD) { int i, j, k; ll temp = 1, t; for (i = 0; i < n; i++) for (j = 0; j < n; j++) a[i][j] %= MOD; for (i = 1; i < n; i++) { for (j = i + 1; j < n; j++) while (a[j][i]) { t = a[i][i] / a[j][i]; for (k = i; k < n; k++) { a[i][k] -= a[j][k] * t; a[i][k] %= MOD; } for (k = i; k < n; k++) swap(a[i][k], a[j][k]); temp = -temp; } temp = temp*a[i][i] % MOD; } return (temp + MOD) % MOD; } ll cal_MST_count(int n, ll MOD) { sort(edge + 1, edge + edgenum + 1, cmp); int pre = edge[1].dis; ll ans = 1; a.init(n); b.init(n); memset(visit, 0, sizeof(visit)); memset(deg, 0, sizeof(deg)); for (int i = 0; i <= n; i++) g[i].clear(); for (int t = 1; t <= edgenum + 1; t++) { if (edge[t].dis != pre || t == edgenum + 1) { for (int i = 1, k; i <= n; i++) if (visit[i]) { k = b.find(i); g[k].push_back(i); visit[i] = 0; } for (int i = 1; i <= n; i++) if (g[i].size()) { memset(p, 0, sizeof(p)); for (int j = 0; j < g[i].size(); j++) for (int k = j + 1, x, y; k < g[i].size(); k++) { x = g[i][j]; y = g[i][k]; p[j][k] = p[k][j] = -deg[x][y]; p[j][j] += deg[x][y]; p[k][k] += deg[x][y]; } ans = ans*DET(p, g[i].size(), MOD) % MOD; for (int j = 0; j < g[i].size(); j++) a.set[g[i][j]] = i; } memset(deg, 0, sizeof(deg)); for (int i = 1; i <= n; i++) { b.set[i] = a.find(i); g[i].clear(); } if (t == edgenum + 1) break; pre = edge[t].dis; } int x = a.find(edge[t].u); int y = a.find(edge[t].v); if (x == y) continue; visit[x] = visit[y] = 1; b.Union(x, y); deg[x][y]++; deg[y][x]++; } if (!edgenum) return 0; for (int i = 2; i <= n; i++) if (b.find(i) != b.find(1)) return 0; return ans; } void init(){ edgenum = 0; } int n, m, u, v, d; ll mod; int main(){ while (cin >> n >> m){ init(); while (m--){ rd(u); rd(v); rd(d); add(u, v, d); } pt(cal_MST_count(n, 31011LL)); putchar(' '); } return 0; }



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  • 原文地址:https://www.cnblogs.com/jzssuanfa/p/7230137.html
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