Bzoj 1016 JSOI2008 minimum spanning tree count Kruskal

Source: Internet
Author: User

Topic: Given an no-show graph, the number of schemes to find the minimum spanning tree

First of all, for a graph of the smallest spanning tree, the number of edges of each edge right is certain

First of all, let's run through the Kruskal, find out the number of occurrences of each edge right on the smallest spanning tree

Then for each of the Benquan that appear on the smallest spanning tree, we start from small to large

For each edge, we enumerate how many of these edge-weighted edges can be added to the minimum spanning tree without forming a ring.

Ans takes this value and then indents all the Unicom blocks that are connected by this edge right.

Note Output 0 when the minimum spanning tree does not exist


#include <map> #include <cstdio> #include <cstring> #include <iostream> #include <algorithm > #define M 1010#define Mo 31011using namespace std;struct edge{int X,y,f;bool operator < (const edge &AMP;Z) const{r Eturn f < z.f;}} Edges[m];int n,m,ans=1,fa[110];int digit[m+100];map<int,int>a;int Find (int x) {if (!fa[x]| | FA[X]==X) return Fa[x]=x;return fa[x]=find (fa[x]);} void Kruskal () {int i,cnt=0;memset (fa,0,sizeof FA); for (i=1;i<=m;i++) {int X=find (edges[i].x), Y=find (EDGES[I].Y); if (x==y) continue;fa[x]=y;cnt++;a[edges[i].f]++;} if (cnt!=n-1) {puts ("0"); exit (0);}} int Deal (int x) {int i,j,y,re=0,cnt=a[edges[x].f];for (y=x;edges[y].f==edges[y+1].f;y++); for (i=0;i<1<<y-x+ 1;i++) if (digit[i]==cnt) {int temp=i;memset (fa,0,sizeof FA); for (j=x;temp;temp>>=1,j++) if (temp&1) {int xx= Find (edges[j].x); int yy=find (EDGES[J].Y); if (Xx==yy) break;fa[xx]=yy;} if (!temp) ++re;} Ans*=re,ans%=mo;memset (fa,0,sizeof FA); for (i=x;i<=y;i++) {int Xx=find (edges[i].x), Yy=find (edges[i].y); if (Xx==yy) continue;fa[xx]=yy;} for (i=y+1;i<=m;i++) Edges[i].x=find (edges[i].x), Edges[i].y=find (EDGES[I].Y); return y;} int main () {int i;for (i=0;i<1<<10;i++) digit[i]=digit[i>>1]+ (i&1); Cin>>n>>m;for (I=1 ; i<=m;i++) scanf ("%d%d%d", &edges[i].x,&edges[i].y,&edges[i].f); sort (edges+1,edges+m+1); Kruskal (); for (i=1;i<=m;i++) if (A[edges[i].f]) i=deal (i); Cout<<ans<<endl;return 0;}


Bzoj 1016 JSOI2008 minimum spanning tree count Kruskal

Contact Us

The content source of this page is from Internet, which doesn't represent Alibaba Cloud's opinion; products and services mentioned on that page don't have any relationship with Alibaba Cloud. If the content of the page makes you feel confusing, please write us an email, we will handle the problem within 5 days after receiving your email.

If you find any instances of plagiarism from the community, please send an email to: info-contact@alibabacloud.com and provide relevant evidence. A staff member will contact you within 5 working days.

A Free Trial That Lets You Build Big!

Start building with 50+ products and up to 12 months usage for Elastic Compute Service

  • Sales Support

    1 on 1 presale consultation

  • After-Sales Support

    24/7 Technical Support 6 Free Tickets per Quarter Faster Response

  • Alibaba Cloud offers highly flexible support services tailored to meet your exact needs.