Bzoj 1016 minimum Spanning tree count

Description

A simple, non-weighted graph is now given. You are not satisfied with finding the smallest spanning tree of the graph, but want to know how many different minimum spanning trees are in the graph. (If there is at least one edge in the two smallest spanning trees, the two minimum spanning trees are different). Because the different minimum spanning tree may be many, you only need to output the scheme number to 31011 modulo.

Input

The first line contains two numbers, N and M, where 1<=n<=100; 1<=m<=1000; Represents the number of nodes and sides of the graph. Each node is numbered with an integer of 1~n. The next M-line, each line contains two integers: A, B, C, the weight of the edge between Node A and B is C, where 1<=c<=1,000,000,000. Data guarantees that no self-back and heavy edges are present. Note: An edge with the same weight will not exceed 10 bars.

Output

How many of the different minimum spanning trees are output. You just need to output a number of 31011 modulo.

Sample Input4 6
1 2 1
1 3 1
1 4 1
2 3 2
2 4 1
3 4 1Sample Output8HINT Source

This question to guess a conclusion--the number of sides of the length of I is certain and the front I small edge they make up and check the set is certain, so you can 2^n Dfs (the same length of Edge <=10).

1#include <algorithm>2#include <cstdio>3#include <cstring>4#include <cstdlib>5 using namespacestd;6 7 #defineMAXN (110)8 #defineMAXM (1010)9 #defineRHL (31011)Ten  One intFATHER[MAXN],SAVE[MAXN],BAC[MAXM],ROAD[MAXM]; A intn,m,tot,ans,sum; - structe{intU,v,w;} EDGE[MAXM]; -  theInlinevoidInit () { for(inti =1; I <= n;++i) father[i] =i;} -  -InlineintFindintA) {if(Father[a]! = a) Father[a] = find (Father[a]);returnfather[a];}  -  +InlineBOOLCMP (E a,e b) {returnA.W <B.W;} -  +InlinevoidMST () A { atSort (edge+1, edge+m+1, CMP); Init (); -     inthave =0, R1,r2,pos; -      for(inti =1; I <= m;++i) -     { -R1 = Find (edge[i].u), r2 =find (EDGE[I].V); -         if(R1! =R2) in         { -FATHER[R1] = R2; ++have ; topos = Lower_bound (bac+1, bac+tot+1, EDGE[I].W)-BAC; +++Road[pos]; -         } the         if(with = = N-1) Break; *     } $     if(With < n-1) printf ("0"), Exit (0);Panax Notoginseng } -  theInlinevoidDfsintAintRintPosintcho) + { A     if(Road[pos] = =cho) the     { +++sum; -         if(Sum = =1) memcpy (Save,father,sizeof(save)); $         return; $     } -     if(A > R)return; -     if(cho+r-a+1<road[pos])return; the     intTEMP[MAXN]; -DFS (A +1, R,pos,cho);Wuyimemcpy (Temp,father,sizeof(temp)); the     intR1 = Find (edge[a].u), r2 =find (EDGE[A].V); -     if(R1! = r2) Father[r1] = R2,dfs (A +1, r,pos,cho+1); Wumemcpy (Father,temp,sizeof(temp)); - } About  $ intMain () - { -Freopen ("1016.in","R", stdin); -Freopen ("1016.out","W", stdout); Ascanf"%d%d",&n,&m); +      for(inti =1; I <= m;++i) the     { -         inta,b,c; $scanf" %d%d%d",&a,&b,&c); theEdge[i] =(E) {a,b,c}; theBac[i] =C; the     } theSort (bac+1, bac+m+1); -tot = unique (bac+1, bac+m+1)-bac-1; in MST (); theInit (); Ans =1; the      for(inti =1; I <=m;) About     { the         intj =i; the          while(J < m && edge[j+1].W = = EDGE[I].W) + +J; thesum =0; +DFS (I,j,lower_bound (bac+1, BAC+TOT,EDGE[I].W)-bac,0); -(Ans *= sum)%=RHL; thememcpy (Father,save,sizeof(save));Bayii = j+1; the     } theprintf"%d", ans); - fclose (stdin); fclose (stdout); -     return 0; the}

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Bzoj 1016 minimum Spanning tree count