Bzoj 1016 [JSOI2008] minimum spanning tree count--matrix-tree theorem

Consider the big plus edge from small to large, and then calculate the number of spanning trees for all the unicom blocks.

Then shrink them to a point and continue adding the next set.

The last multiplication principle can be.

It's disgusting, you write.

#include <queue> #include <cmath> #include <vector> #include <cstdio> #include <cstring># Include <iostream> #include <algorithm>using namespace std;  #define F (I,J,K) for (int. i=j;i<=k;++i) #define D (I,J,K) for (int i=j;i>=k;--i) #define MAXN 1005#define EPS 1e-6const int md=31011;vector <int> v,to[maxn];queue <int> q;struct edge{int u,v,w;} A[maxn];int n,m,fa[maxn];int b[maxn][maxn],inv[maxn];int vcnt,du[maxn],list[maxn],vis[maxn];bool cmp (Edge X,Edge y) { return X.W&LT;Y.W;} int GF (int k) {if (fa[k]==k) return k; else return FA[K]=GF (Fa[k]);}    int gauss (int n) {f (i,1,n) F (j,1,n) b[i][j]%=md;    int ret=1;  for (int i=1;i<n;++i) {to (int j=i+1;j<n;++j) while (B[j][i]) {int                T=b[i][i]/b[j][i];                for (int k=i;k<n;++k) b[i][k]= (B[I][K]-B[J][K]*T+MD)%md;                for (int k=i;k<n;++k) swap (b[i][k],b[j][k]); REt=-ret;        } if (b[i][i]==0) return 0;    RET=RET*B[I][I]%MD; } return ABS ((RET+MD)%MD);}    int main () {scanf ("%d%d", &n,&m);    F (i,1,n) fa[i]=i;    F (i,1,m) {scanf ("%d%d%d", &AMP;A[I].U,&AMP;A[I].V,&AMP;A[I].W);}    Sort (a+1,a+m+1,cmp);    int Now=1,ans=1;        while (now<=m) {int l=now,r=now;        vcnt=0;        memset (du,0,sizeof du);        F (I,1,n) to[i].clear ();        while (A[R+1].W==A[R].W) r++;        now=r+1;            F (i,l,r) {int FL=GF (A[I].U), FR=GF (A[I].V);            To[fl].push_back (FR);            To[fr].push_back (FL);        if (FL!=FR) du[fl]++,du[fr]++;        } memset (vis,0,sizeof vis);                F (I,1,n) if (Du[i]&&!vis[i]) {v.clear ();                memset (b,0,sizeof b);                memset (inv,0,sizeof inv);                Q.push (i); inv[i]=1;vis[i]=1; while (!q.empty ()) {int X=q.front (); V.push_baCK (x); Q.pop (); for (int j=0;j<to[x].size (); ++j) if (!vis[to[x][j]]) Q.push (To[x][j])                , Inv[to[x][j]]=1,vis[to[x][j]]=1;                } for (int j=0;j<v.size (); ++j) list[v[j]]=j+1; for (int j=0;j<v.size (), ++j) for (int k=0;k<to[v[j]].size (); ++k) if (inv[t O[v[j]][k]]) {b[list[v[j]]][list[v[j]]]++,b[list[v[j]]][list[to[v[j]][k                        ]]]--;                } Ans*=gauss (V.size ());            ANS%=MD;            } F (i,l,r) {int FL=GF (A[I].U), FR=GF (A[I].V);        if (FL!=FR) {fa[fl]=fr;}    }} int cnt=0;        F (I,1,n) if (fa[i]==i) {cnt++;    if (cnt==2) {printf ("0\n"); return 0;} } printf ("%d\n", ans);}

　　

Bzoj 1016 [JSOI2008] minimum spanning tree count--matrix-tree theorem