BZOJ1016: [JSOI2008] minimum spanning tree count

Surface

Transmission Door

Sol

The nature of the minimum spanning tree:

• For each \ (mst\), the same number of edges is used for each of the edge weights
• All (mst\) edge \ (≤w\) sides of the graph are the same connectivity

Then the problem is enumerated without a weight to select those sides, if the number of connected and the same as the original statistics
The last multiplication principle can

If the same Benquan too much, you can only use the matrix tree theorem
But I'm too good to be.

# include <bits/stdc++.h># define RG Register# define IL inline# define Fill (A, B) memset (A, B, sizeof (a))# define File (a) freopen (a ".", "R", stdin), Freopen (a ". Out", "w", stdout)using namespaceStdtypedef Long LongllConst intZsy (31011);Const int_( the);Const int__(1005); ILintInput () {RGintx =0, z =1; RgCharc = GetChar (); for(; c <' 0 '|| C >' 9 '; c = GetChar ()) z = c = ='-'?-1:1; for(; C >=' 0 '&& C <=' 9 '; c = GetChar ()) x = (x <<1) + (x <<3) + (c ^ -);returnx * z;}intN, M, fa[_], ans =1, o[__], Len, l[__], r[__], id[__], cnt[__], tmp[_];structedge{intU, V, W; ILBOOL operator< (RG Edge B)Const{returnW < B.W; }} Edge[__];ilintFind (RGintx) {returnx = = Fa[x]? X:FA[X] = Find (Fa[x]);} ILintCount (RGintx) {RGintRET =0; for(; x; x-= x &-X) ++ret;returnRET;}intMain (RGintARGC, RGChar* argv[]) {n = input (), M = input (); for(RGinti =1; I <= N; ++i) Fa[i] = i; for(RGinti =1; I <= m;        ++i) {Edge[i] = (edge) {input (), input (), input ()}; O[i] = EDGE[I].W, l[i] = m, r[i] =1; } Sort (Edge +1, Edge + M +1); Sort (o +1, O + M +1), Len = unique (o +1, O + M +1)-O-1; RgintGG =0; for(RGinti =1; I <= m; ++i) {RGintFX = Find (edge[i].u), FY = find (EDGE[I].V); Id[i] = Lower_bound (o +1, O + len +1, EDGE[I].W)-O; L[id[i]] = min (L[id[i]], i), r[id[i]] = max (R[id[i]], i);if(Find (FX) = = Find (FY))Continue;    FA[FX] = FY, ++cnt[id[i]], ++gg; }if(GG! = n-1)returnPuts"0"),0; for(RGinti =1; I <= N; ++i) Fa[i] = i; for(RGinti =1; I <= Len; ++i) {RGinttot =0; for(RGintj =1; J <= N; ++J) Tmp[j] = Fa[j]; for(RGintj =0, S =1<< (R[i]-l[i] +1); J < S; ++J) {if(Count (j)! = Cnt[i])Continue; RgintFLG =0; for(RGintK =1; K <= N; ++K) Fa[k] = tmp[k]; for(RGintK =0; K <= R[i]-l[i]; ++K)if((1<< k) & j) {RGintFX = Find (Edge[k + l[i]].u), FY = find (edge[k + l[i]].v);if(FX = = FY) {FLG =1; Break;                } Fa[fx] = FY; }if(!FLG) ++tot; } ans = ans * tot% Zsy; for(RGintj =1; J <= N; ++J) Fa[j] = Tmp[j]; for(RGintj = L[i]; J <= R[i]; ++J) {RGintFX = Find (edge[j].u), FY = find (EDGE[J].V);if(FX! = FY) fa[fx] = FY; }} printf ("%d\n", ans);return 0;}

BZOJ1016: [JSOI2008] minimum spanning tree count