hdu 5556 Land of Farms

Land of Farms

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 675    Accepted Submission(s): 220

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

2015ACM/ICPC亚洲区合肥站-重现赛（感谢中科大）

 

Recommend

wange2014   |   We have carefully selected several similar problems for you:   6032  6031  6030  6029  6028 

 

题意： 一个N*M的矩阵，其中“.”代表空地，“0-9”代表古代建筑，相邻的相同古代建筑要么不建要么同时建，即可以看成一个点，每个空地也是一个点，然后相邻的两个点不能同时建，问最多能建几个点。

思路：当时我还没做过最大团问题。我先把联通块缩点然后把非相邻的两个点建边，然后就想不到怎么搞了。我只知道我这建的图的的最大完全子图是答案，但是不会求。之后找了题解就看到了最大团这个概念。最大团就是最大完全子图。然后我把模板套上去就好了。。。。。这里还有个定理：最大独立集合=补图的最大团=节点数-最大匹配数。下面给代码：

#include<bits/stdc++.h> 
using namespace std;
typedef long long LL;
const int maxn=15;
char s[maxn][maxn];
int n,m,vis[maxn][maxn],steparr[4][2]={-1,0,1,0,0,1,0,-1},vis2[maxn*maxn][maxn*maxn],num;
const int N=105;
bool G[N][N];
int Max[N],Alt[N][N],ans;
void dfsblock(int x,int y,char mark){
	if(x<0||x>=n||y<0||y>=m||vis[x][y]||s[x][y]!=mark)
		return;
	vis[x][y]=num;
	for(int i=0;i<4;i++){
		int nextx=x+steparr[i][0];
		int nexty=y+steparr[i][1];
		dfsblock(nextx,nexty,mark);
	}
}
void dfsedge(int x,int y){
	if(x<0||x>=n||y<0||y>=m||vis2[x][y])
		return;
	vis2[x][y]=1;
	for(int i=0;i<4;i++){
		int nextx=x+steparr[i][0];
		int nexty=y+steparr[i][1];
		G[vis[x][y]][vis[nextx][nexty]]=1;
		dfsedge(nextx,nexty);
	}
}

bool DFS(int cur,int tot){
    if(!cur){
        if(tot>ans){ans=tot;return 1;}
        return 0;
    }
    for(int i=1;i<=cur;i++){
        if(cur-i+tot+1<=ans)return 0;
        int u=Alt[tot][i],nxt=0;
        if(Max[u]+tot<=ans)return 0;
        for(int j=i+1;j<=cur;j++)
            if(G[u][Alt[tot][j]])Alt[tot+1][++nxt]=Alt[tot][j];
        if(DFS(nxt,tot+1))return 1;
    }
        return 0;
    }
int MaxClique(){
    ans=0,memset(Max,0,sizeof(Max));
    for(int i=num-1;i;i--){
        int cur=0;
        for(int j=i+1;j<num;j++)
            if(G[i][j])Alt[1][++cur]=j;
        DFS(cur,1);
        Max[i]=ans;
    }
    return ans;
}
int main(){
	int t;
	scanf("%d",&t);
	for(int tcase=1;tcase<=t;tcase++){
		memset(vis,0,sizeof(vis));
		memset(G,0,sizeof(G));
		memset(vis2,0,sizeof(vis2));
		ans=0;
		num=1;
		scanf("%d%d",&n,&m);
		for(int i=0;i<n;i++)
			scanf("%s",s[i]);
		for(int i=0;i<n;i++){
			for(int j=0;j<m;j++){
				if(vis[i][j])
					continue;
				if(s[i][j]=='.')
					vis[i][j]=num++;
				else{
					dfsblock(i,j,s[i][j]);
					num++;
				}
			}
		}	
		dfsedge(0,0);
		for(int i=1;i<num;i++){
			for(int j=1;j<num;j++){
				G[i][j]=!G[i][j];
			}
		}
		MaxClique();
		printf("Case #%d: %d\n",tcase,ans);
	}
}

#include<bits/stdc++.h>
using namespace std;
#define N 102  
int mx;//最大团数(要初始化为0)  
int x[N],tuan[N];  
int can[N][N];//can[i]表示在已经确定了经选定的i个点必须在最大团内的前提下还有可能被加进最大团的结点集合  
int num[N];//num[i]表示由结点i到结点n构成的最大团的结点数  
bool g[N][N];//邻接矩阵(从1开始)  
int n,m;  
bool dfs(int tot,int cnt)
{  
    int i,j,k;  
    if(tot == 0)
	{  
        if(cnt > mx)
		{  
            mx = cnt;  
            for(i=0;i<mx;i++)
			{  
                tuan[i] = x[i];  
            }  
            return true;  
        }  
        return false;  
    }  
    for(i=0;i<tot;i++)
	{  
        if(cnt + (tot-i) <= mx)return false;  
        if(cnt + num[can[cnt][i]] <= mx)return false;  
        k = 0;  
        x[cnt] = can[cnt][i];  
        for(j=i+1;j<tot;j++)
		{  
            if(g[can[cnt][i]][can[cnt][j]])
			{  
                can[cnt+1][k++] = can[cnt][j];  
            }  
        }  
        if(dfs(k,cnt+1))return false;  
    }  
    return false;  
}  
void MaxTuan()
{  
    int i,j,k;  
    mx = 1;  
    for(i=n;i>=1;i--)
	{  
        k = 0;  
        x[0] = i;  
        for(j=i+1;j<=n;j++)
		{  
            if(g[i][j])
			{  
                can[1][k++] = j;  
            }  
        }  
        dfs(k,1);  
        num[i] = mx;  
    }  
}  
const int maxn=20;
struct Node
{
	int x,y;
	Node(int _x=0,int _y=0):x(_x),y(_y){}
};
int id,dir[4][2]={-1,0,1,0,0,-1,0,1},ID[maxn][maxn];
char mp[maxn][maxn];
bool over(int x,int y)
{
	if(x<1||x>n||y<1||y>m)	return true;
	return false;
}
void getid(int x,int y)
{
	if(ID[x][y]) return;
	if(mp[x][y]=='.') { ID[x][y]=id++; return; }
	else ID[x][y]=id++;
	queue<Node>	Q;
	Q.push(Node(x,y));
	while(!Q.empty())
	{
		Node head=Q.front();
		Q.pop();
		for(int i=0;i<4;i++)
		{
			int tx=head.x+dir[i][0],ty=head.y+dir[i][1];
			if(over(tx,ty)||ID[tx][ty]||mp[tx][ty]-mp[x][y]) continue;
			ID[tx][ty]=ID[x][y];
			Q.push(Node(tx,ty));
		}
	}
}
void init()
{
	memset(ID,0,sizeof(ID));
	memset(g,0,sizeof(g)); 
	id=1; 
	mx=0;
}
int main()
{  
    int T,cas=1;
    scanf("%d",&T);
	while(T--)
	{  
		init();
		scanf("%d%d",&n,&m);
		for(int i=1;i<=n;i++) scanf("%s",mp[i]+1);
		for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) getid(i,j);
        for(int i=1;i<=n;i++)
        {
			for(int j=1;j<=m;j++)
			{
				for(int k=0;k<4;k++)
				{
					int x=i,y=j;
					int tx=x+dir[k][0],ty=y+dir[k][1];
					if(over(tx,ty)) continue;
					if(ID[x][y]!=ID[tx][ty]) g[ID[x][y]][ID[tx][ty]]=g[ID[tx][ty]][ID[x][y]]=true;
				}
			}
		}
		n=id-1;
        for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) g[i][j]=!g[i][j];
        MaxTuan();  
        printf("Case #%d: %d\n",cas++,mx);  
    }  
    return 0;  
}