HDU 5727 Necklace two points + match

This is from the Shandong University giant to learn from the practice

The total number of black balls in the enumeration is 8!, and then eight white balls are available in position with two black balls left and right there is no relationship map

This is the exact words, the specific point, each generation of the ring row, only the non-impact of the connected edge

Finally: Pay attention to a little, n number of ring row is (n-1)!

#include <stdio.h>#include<iostream>#include<algorithm>#include<string.h>#include<vector>#include<map>using namespaceStd;typedefLong LongLL;Const intinf=0x3f3f3f3f;Const intn=2e5+5;intpos[Ten],n,m,match[Ten],ret;BOOLvis[Ten],used[Ten],g[Ten][Ten],mp[Ten][Ten];BOOLDfsintu) {   for(intv=1; v<=n;++v) {    if(!g[u][v]| | USED[V])Continue; USED[V]=true; if(match[v]==-1||DFS (Match[v])) {Match[v]=T; return true; }  }  return false;}voidsolve () {memset (match,-1,sizeof(match)); Memset (g,false,sizeof(g));  for(intI=2; i<=n;++i) {     for(intj=1; j<=n;++j) {      if(mp[pos[i]][j]| | mp[pos[i-1]][j])Continue; G[I][J]=true; }  }   for(intI=1; i<=n;++i) {    if(mp[pos[1]][i]| |Mp[pos[n]][i])Continue; g[1][i]=true; }  intans=0;  for(intI=1; i<=n;++i) {memset (used,false,sizeof(used)); if(Dfs (i)) + +ans; } ret=min (ret,n-ans);}void Get(intx) {  if(ret==0)return; if(x==n+1) {solve ();return;}  for(intI=1; i<=n;++i) {    if(Vis[i])Continue; POS[X]=i; Vis[i]=true; Get(x+1); Vis[i]=false; }}intMain () {vis[1]=true;p os[1]=1;  while(~SCANF ("%d%d",&n,&m)) {     if(n==0) {printf ("0\n"); Continue; } memset (MP,0,sizeof(MP));  while(m--){      intu,v; scanf ("%d%d",&u,&v); Mp[v][u]=true; } ret=INF; Get(2); printf ("%d\n", ret); }  return 0;}

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HDU 5727 Necklace two points + match