【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('
')
#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();
}