4185 create a maximum matching parity chart for oil skimming

Question:

Count the logarithm of '#' of adjacent (upper and lower left.

Solution:

Similar to the hdu1507 Uncle Tom's inherited land *, you need to use a parity graph. That is, the row + column is an odd number of X sets, and the column is occasionally used as a y set. It is a '#' Even edge. Finally, find the maximum match.

The data is a little big, and an error occurs when creating a graph directly (I tried it ). The vertex numbers are displayed in the order.

  1 #include<iostream>  2 #include<cstdio>  3 #include<cstring>  4 #include<queue>  5 #include<vector>  6 #include<algorithm>  7 using namespace std;  8 const int N=36005,INF=0x3f3f3f3f;  9 int cx[N],cy[N],dx[N],dy[N]; 10 bool bmask[N]; 11 vector<int > bmap[N]; 12 int nx,ny,dis,ans; 13 bool searchpath() 14 { 15     queue<int> q; 16     dis=INF; 17     memset(dx,-1,sizeof(dx)); 18     memset(dy,-1,sizeof(dy)); 19     for(int i=1;i<=nx;i++) 20     { 21         if(cx[i]==-1){ q.push(i); dx[i]=0; } 22         while(!q.empty()) 23         { 24             int u=q.front(); q.pop(); 25             if(dx[u]>dis) break; 26             for(int i=0;i<bmap[u].size();i++) 27             { 28                 int v=bmap[u][i]; 29                 if(dy[v]==-1) 30                 { 31                     dy[v]= dx[u] + 1; 32                     if(cy[v]==-1) dis=dy[v]; 33                     else 34                     { 35                         dx[cy[v]]= dy[v]+1; 36                         q.push(cy[v]); 37                     } 38                 } 39             } 40         } 41     } 42     return dis!=INF; 43 } 44 int findpath(int u) 45 { 46     for(int i=0;i<bmap[u].size();i++) 47     { 48         int v=bmap[u][i]; 49         if(!bmask[v]&&dy[v]==dx[u]+1) 50         { 51             bmask[v]=1; 52             if(cy[v]!=-1&&dy[v]==dis) continue; 53             if(cy[v]==-1||findpath(cy[v])) 54             { 55                 cy[v]=u; cx[u]=v; 56                 return 1; 57             } 58         } 59     } 60     return 0; 61 } 62 void maxmatch() 63 { 64     ans=0; 65     memset(cx,-1,sizeof(cx)); 66     memset(cy,-1,sizeof(cy)); 67     while(searchpath()) 68     { 69         memset(bmask,0,sizeof(bmask)); 70         for(int i=1;i<=nx;i++) 71             if(cx[i]==-1) ans+=findpath(i); 72     } 73 } 74 void init() 75 { 76     for(int i=0;i<=N;i++) bmap[i].clear(); 77 } 78 char s[605][605]; 79 int Map[605][605]; 80 int main() 81 { 82     //freopen("test.txt","r",stdin); 83     int cas,n,i,j,k=1,a,b,t=1; 84     char ch; 85     scanf("%d",&cas); 86     while(cas--) 87     { 88         scanf("%d",&n); 89         t=0; 90         for(i=1;i<=n;i++){ 91             getchar(); 92             for(j=1;j<=n;j++) 93             { 94                 scanf("%c",&s[i][j]); 95                 if(s[i][j]==‘#‘) Map[i][j]=++t; 96             } 97         } 98         init(); 99         for(i=1;i<=n;i++)100         {101             for(j=1;j<=n;j++)102             {103                 if(s[i][j]==‘#‘&&((i+j)%2))104                 {105                     a=Map[i][j];106                     if(i>1&&s[i-1][j]==‘#‘){107                         b=Map[i-1][j];108                         bmap[a].push_back(b);109                     }110                     if(i<n&&s[i+1][j]==‘#‘){111                         b=Map[i+1][j];112                         bmap[a].push_back(b);113                     }114                     if(j>1&&s[i][j-1]==‘#‘){115                         b=Map[i][j-1];116                         bmap[a].push_back(b);117                     }118                     if(j<n&&s[i][j+1]==‘#‘){119                         b=Map[i][j+1];120                         bmap[a].push_back(b);121                     }122                 }123             }124         }125         nx=t;126         maxmatch();127         printf("Case %d: %d\n",k++,ans);128     }129     return 0;130 }

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4185 create a maximum matching parity chart for oil skimming