Oil collection (for Unicom area) more than 2018 School Winter Camp (dfs+ two matching)

Topic:

Links: Https://www.nowcoder.com/acm/contest/76/A
Source: Niu Ke Net

With the issue of oil spills at sea, a new lucrative industry is being created, which is the oil-skimming industry. Today, a large amount of oil floating in the Gulf of Mexico attracts a lot of businessmen's attention. These businessmen have a special kind of plane, can be a scoop over the entire sea 20 meters by 10 meters such a large rectangle. (adjacent or left and right adjacent to the lattice, can not be inclined to come) of course, this requires a scoop of the past is all oil, if there is oil in a ladle there is water, it is meaningless, the resources are completely unusable. Now, businessmen want to know how much oil he can get in this area.

The map is a NXN network, each of which represents a square area of 10mx10m, each of which is marked with oil or water.

Input Description:

Test input contains multiple test data
The first line of the test data gives the number of test data T (t<75)
Each test sample uses the number n (n<50) to represent the size of the map area, followed by n lines with n characters in each row, where the symbol '. ' Indicates the surface of the sea, symbol ' # ' represents the oil surface.

Output Description:

The output format is "Case X:m" (X starting from 1) and M is the maximum amount of oil a trader can get.

Example 1 input

16.......##..........#. #.. #.. ##......

Output

Case 1:3



Test instructions: The Chinese question does not explain



Idea: Way1:dfs is ok ah,,, so simple topic do not know why at that time did not out
     Way2: two-point matching

Code Listing 1:

#include <iostream>#include<cstdio>#include<cstring>#include<cmath>using namespacestd;intn,m;Charmp[106][106];intdx[4]={0,1,0,-1};intdy[4]={1,0,-1,0};intAns1,ans2;voidDFS (intIintj) {    intNx,ny; if(i<0|| i>=m| | j<0|| j>=n| | mp[i][j]=='.')        return ; if((i+j)%2==0) ans1++; Elseans2++; MP[I][J]='.';  for(intzz=0;zz<4; zz++) {NX=i+Dx[zz]; NY=j+Dy[zz];    DFS (Nx,ny); }    return ;}intMain () {inti,j,res=0; intT; CIN>>T;  for(into=1; o<=t;o++) {cin>>m; N=m;        GetChar ();  for(i=0; i<m;i++)        {             for(j=0; j<n;j++) {scanf ("%c",&Mp[i][j]);        } getchar (); }         for(i=0; i<m;i++)             for(j=0; j<n;j++)        {            if(mp[i][j]=='#') {ans1=0; Ans2=0;                DFS (I,J); Res+=min (ans1,ans2); }} printf ("Case %d:%d\n", O,res); }    return 0;}

Code 2: (The idea is simply not to write it yourself)

#include <iostream>#include<queue>#include<algorithm>#include<string.h>#include<math.h>#include<map>using namespacestd;Const intmaxn= -;intgirl[maxn],used[maxn],line[maxn][maxn],path[ -][ -],TEMP1,TEMP2;Chara[ -][ -];BOOLFindintx) {     for(intI=1; i<temp2;i++)    if(Line[x][i]&&!used[i])//x and I have a relationship{Used[i]=1; if(girl[i]==0|| Find (Girl[i]))//no master or can make a position;{Girl[i]=x; return true; }    }    return false;}intMain () {intT,n,ans,tt; CIN>>T; TT=1;  while(t--) {ans=0; Temp1=temp2=1; memset (line,0,sizeof(line)); memset (Girl,0,sizeof(girl)); memset (Path,0,sizeof(path)); CIN>>N;  for(intI=1; i<=n;i++)             for(intj=1; j<=n;j++) {cin>>A[i][j]; if((i+j)%2==0) Path[i][j]=temp1++; ElsePath[i][j]=temp2++; }         for(intI=1; i<=n;i++)             for(intj=1; j<=n;j++)            {                if((i+j)%2==1&&a[i][j]=='#')                {                    if(path[i-1][j]>=1&&a[i-1][j]=='#') Line[path[i-1][j]][path[i][j]]=1; if(path[i+1][j]>=1&&a[i+1][j]=='#') Line[path[i+1][j]][path[i][j]]=1; if(path[i][j-1]>=1&&a[i][j-1]=='#') Line[path[i][j-1]][path[i][j]]=1; if(path[i][j+1]>=1&&a[i][j+1]=='#') Line[path[i][j+1]][path[i][j]]=1; }            }         for(intI=1; i<temp1;i++) {memset (used,0,sizeof(used)); if(Find (i)) ans++; } cout<<" Case"<<tt++<<": "<<ans<<Endl; }}

Simple questions

Oil collection (for Unicom area) more than 2018 School Winter camp (dfs+ match)