Land of Farms
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 539 Accepted Submission(s): 177
Problem Description
Farmer John and his brothers have found a new land. They are so excited and decide to build new farms on the land. The land is a rectangle and consists of N×M grids. A farm consists of one or more connected grids. Two grids are adjacent if they share a common border, i.e. their Manhattan distance is exactly 1. In a farm, two grids are considered connected if there exist a series of adjacent grids, which also belong to that farm, between them.
Farmer John wants to build as many farms as possible on the new land. It is required that any two farms should not be adjacent. Otherwise, sheep from different farms would fight on the border. This should be an easy task until several ancient farms are discovered.
Each of the ancient farms also consists of one or more connected grids. Due to the respect to the ancient farmers, Farmer John do not want to divide any ancient farm. If a grid from an ancient farm is selected in a new farm, other grids from the ancient farm should also be selected in the new farm. Note that the ancient farms may be adjacent, because ancient sheep do not fight each other.
The problem is a little complicated now. Can you help Farmer John to find a plan with the maximum number of farms?
Input
The first line of input contains a number T indicating the number of test cases (T≤200).
Each test case starts with a line containing two integers N and M, indicating the size of the land. Each of the following N lines contains M characters, describing the map of the land (1≤N,M≤10). A grid of an ancient farm is indicated by a single digit (0-9). Grids with the same digit belong to the same ancient farm. Other grids are denoted with a single character “.”. It is guaranteed that all test cases are valid.
Output
For each test case, output a single line consisting of “Case #X: Y”. X is the test case number starting from 1. Y is the maximum number of new farms.
Sample Input
3
3 4
..3.
023.
.211
2 3
...
...
4 4
1111
1..1
1991
1111
Sample Output
Case #1: 4
Case #2: 3
Case #3: 1
Source
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题意： 有一个农场大小为n*m 里面有一些古老的农田，你现在需要新建一些农田，要求新建的农田之间不能相连，古老的农田不可拆分，如果你选择了一块土地（原为古老的农田）建立新农田则需要把该块古老的农田全部包含。
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <iostream>
#include <cmath>
#include <map>
#include <bitset>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <set>
#define MM(a,b) memset(a,b,sizeof(a));
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
#define CT continue
#define SC scanf
char f[15][15];
int anc[16],flag[14],mp[12][12],cnta,ans,res;
int dx[]={0,0,1,-1},dy[]={1,-1,0,0},match[105],used[105];
vector<int> G[205];
int cas,r,l;
bitset<15> sta;
void add_edge(int u,int v)
{
G[u].push_back(v);
G[v].push_back(u);
//if(sta==0) cout<<"u: "<<u<<" v:"<<v<<"\n";
}
bool dfs(int u)
{
used[u]=1;
for(int i=0;i<G[u].size();i++){
int v=G[u][i],w=match[v];
if(w<0||(!used[w]&&dfs(w))){
match[u]=v;
match[v]=u;
return true;
}
}
return false;
}
int bi_mactch()
{
int res=0;
MM(match,-1);
for(int i=1;i<=r;i++) for(int j=1;j<=l;j++){
int k=(i-1)*l+j;
if(match[k]<0){
MM(used,0);
if(dfs(k)) res++;
}
}
return res;
}
int par[15];
int findr(int u)
{
if(par[u]!=u)
par[u]=findr(par[u]);
return par[u];
}
void unite(int u,int v)
{
int ru=findr(u),rv=findr(v);
if(ru!=rv) par[ru]=rv;
}
int num=0;
void sear(int i,int j)
{
if(f[i][j]>=‘0‘&&f[i][j]<=‘9‘){
int k=f[i][j]-‘0‘;
if(!flag[k]) return;
mp[i][j]=-1;
for(int d=0;d<4;d++){
int tx=i+dx[d],ty=j+dy[d];
if(f[tx][ty]>=‘0‘&&f[tx][ty]<=‘9‘){
int k2=f[tx][ty]-‘0‘;
if(flag[k2]) unite(k,k2);
}
else if(f[tx][ty]==‘.‘) mp[tx][ty]=-1;
}
}
}
void bgraph()
{
for(int i=1;i<=110;i++) G[i].clear();
for(int i=1;i<=r;i++)
for(int j=1;j<=l;j++)
if(f[i][j]==‘.‘&&!mp[i][j]){
num++;
int k=(i-1)*l+j;
//if(sta==0) cout<<"kk:"<<k<<"\n";
if(f[i-1][j]==‘.‘&&!mp[i-1][j]) add_edge(k,k-l);
if(f[i][j-1]==‘.‘&&!mp[i][j-1]) add_edge(k,k-1);
}
}
void solve()
{
ans=0;
for(int k=0;k<=(1<<cnta)-1;k++){
res=0;MM(mp,0);MM(flag,0);
sta=k;
for(int i=0;i<cnta;i++) if(sta[i]) {
int w=anc[i+1];flag[w]=1;
par[w]=w;
}
for(int i=1;i<=r;i++) for(int j=1;j<=l;j++) sear(i,j);
for(int i=1;i<=cnta;i++)
if(flag[anc[i]]&&par[anc[i]]==anc[i]) res++;
num=0;bgraph();
ans=max(ans,res+num-bi_mactch());
}
}
int main()
{
SC("%d",&cas);
int kk=0;
while(cas--){
SC("%d%d",&r,&l);
MM(flag,0);
cnta=0;
for(int i=1;i<=r;i++) {
SC("%s",f[i]+1);
for(int j=1;j<=l;j++)
if(f[i][j]!=‘.‘){
int k=f[i][j]-‘0‘;
if(!flag[k]){
flag[k]=1;
anc[++cnta]=k;
}
}
}
solve();
printf("Case #%d: %d\n",++kk,ans);
}
return 0;
}
分析：
1.先2^10=10^3暴力枚举选择的古代的田，选择了后，在用并查集维护一下，能产生的牧田数。
2.删除选择的古代的田周围的普通田地，再用下网格的二分图，考虑不想邻的连接一条边，那么最后显然成成了，求这样一个最大团
3.二分图的最大团=补图的最大点独立集合=顶点数-最大匹配数