【BZOJ】1016: [JSOI2008]最小生成树计数

题解

考虑kruskal
我们都是从边权最小的边开始取，然后连在一起

那我们选出边权最小的一堆边，然后这个图就分成了很多联通块，把每个联通块内部用矩阵树定理算一下生成树个数，再把联通块缩成一个大点，重复取下一个边权的边进行操作

好想然而不是很好写= =写起来感觉有点麻烦

模数非质数，用long double水一下过掉了

代码

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <vector>
#include <set>
#include <cmath>
#include <bitset>
#include <queue>
#define enter putchar('\n')
#define space putchar(' ')
//#define ivorysi
#define pb push_back
#define mo 974711
#define pii pair<int,int>
#define mp make_pair
#define fi first
#define se second
#define MAXN 200005
#define eps 1e-12
using namespace std;
typedef long long int64;
typedef long double db;
template<class T>
void read(T &res) {
res = 0;char c = getchar();T f = 1;
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 - '0' + c;
c = getchar();
}
res = res * f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) out(x / 10);
putchar('0' + x % 10);
}
int N,M;
struct node {
int u,v,c;
friend bool operator < (const node &a,const node &b) {
return a.c < b.c;
}
}E[1005];
int fa[105],id[105],f[105][105],L[105],tot,D[105];
bool vis[105];
db g[105][105];
int64 ans = 1;
int getfa(int x) {
return x == fa[x] ? x : fa[x] = getfa(fa[x]);
}
db Guass(int n) {
db res = 1;
for(int i = 2 ; i <= n ; ++i) {
int l = i;
for(int j = i + 1 ; j <= n ; ++j) {
if(fabs(g[j][i]) > fabs(g[l][i])) l = j;
}
if(i != l) {
res = -res;
for(int j = i ; j <= n ; ++j) {
swap(g[i][j],g[l][j]);
}
}
if(fabs(g[i][i]) == 0) return 0;
for(int j = i + 1 ; j <= n ; ++j) {
db t = g[j][i] / g[i][i];
for(int k = i ; k <= n ; ++k) {
g[j][k] -= g[i][k] * t;
}
}
}
for(int k = 2 ; k <= n ; ++k) {
res = res * g[k][k];
}
return res;
}
void dfs(int u,int n) {
vis[u] = 1;
L[++tot] = u;
D[u] = tot;
for(int i = 1 ; i <= n ; ++i) {
if(f[u][i]) {
if(!vis[i]) dfs(i,n);
}
}
}
void Process(int l,int r) {
memset(id,0,sizeof(id));
memset(f,0,sizeof(f));
memset(vis,0,sizeof(vis));
int cnt = 0;
for(int i = 1 ; i <= N ; ++i) {
if(!id[getfa(i)]) {
id[getfa(i)] = ++cnt;
}
}
for(int i = l ; i <= r ; ++i) {
int s = getfa(E[i].u),t = getfa(E[i].v);
if(s == t) continue;
f[id[s]][id[t]]++; f[id[t]][id[s]]++;
}
for(int i = 1 ; i <= cnt ; ++i) {
if(!vis[i]) {
tot = 0;
dfs(i,cnt);
if(tot == 1) continue;
memset(g,0,sizeof(g));
for(int j = 1 ; j <= tot ; ++j) {
int u = L[j];
for(int k = 1 ; k <= cnt ; ++k) {
if(f[u][k]) {
g[j][j] += f[u][k];
g[j][D[k]] -= f[u][k];
}
}
}
ans = ans * ((int64)(Guass(tot) + 0.5) % 31011) % 31011;
}
}
for(int i = l ; i <= r ; ++i) {
fa[getfa(E[i].v)] = getfa(E[i].u);
}
}
void Solve() {
read(N);read(M);
for(int i = 1 ; i <= N ; ++i) fa[i] = i;
int u,v,c;
for(int i = 1 ; i <= M ; ++i) {
read(E[i].u);read(E[i].v);read(E[i].c);
fa[getfa(E[i].u)] = getfa(E[i].v);
}
for(int i = 2 ; i <= N ; ++i) {
if(getfa(i) != getfa(i - 1)) {
puts("0");
return;
}
}
for(int i = 1 ; i <= N ; ++i) fa[i] = i;
sort(E + 1,E + M + 1);
v = 0;int st = 0;
bool flag = 0;
for(int i = 1 ; i <= M ; ++i) {
if(E[i].c != v) {
if(st != 0) Process(st,i - 1);
st = i;
v = E[i].c;
flag = 1;
for(int j = 2 ; j <= N ; ++j) {
if(getfa(j) != getfa(j - 1)) {flag = 0;break;}
}
if(flag) break;
}
}
if(!flag) Process(st,M);
out(ans);enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
}