Manhattan Wiring

Time Limit: 5000MS	?	Memory Limit: 65536K
Total Submissions: 1110	?	Accepted: 634

Description

There is a rectangular area containing n × m cells. Two cells are marked with “2”, and another two with “3”. Some cells are occupied by obstacles. You should connect the two “2”s and also the two “3”s with non-intersecting lines. Lines can run only vertically or horizontally connecting centers of cells without obstacles.

Lines cannot run on a cell with an obstacle. Only one line can run on a cell at most once. Hence, a line cannot intersect with the other line, nor with itself. Under these constraints, the total length of the two lines should be minimized. The length of a line is defined as the number of cell borders it passes. In particular, a line connecting cells sharing their border has length 1.

Fig. 1(a) shows an example setting. Fig. 1(b) shows two lines satisfying the constraints above with minimum total length 18.

Figure 1: An example of setting and its solution

Input

The input consists of multiple datasets, each in the following format.

n	m
row1
…
rown

n is the number of rows which satisfies 2 ≤ n ≤ 9. m is the number of columns which satisfies 2 ≤ m ≤ 9. Each row_i is a sequence of m digits separated by a space. The digits mean the following.

0: Empty

1: Occupied by an obstacle

2: Marked with “2”

3: Marked with “3”

The end of the input is indicated with a line containing two zeros separated by a space.

Output

For each dataset, one line containing the minimum total length of the two lines should be output. If there is no pair of lines satisfying the requirement, answer “0” instead. No other characters should be contained in the output.

Sample Input

5 5 0 0 0 0 0 0 0 0 3 0 2 0 2 0 0 1 0 1 1 1 0 0 0 0 3 2 3 2 2 0 0 3 3 6 5 2 0 0 0 0 0 3 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 2 3 0 5 9 0 0 0 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 2 0 0 0 0 0 2 0 0 0 0 0 3 0 0 0 0 0 0 0 0 0 0 0 0 0 9 9 3 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 3 9 9 0 0 0 1 0 0 0 0 0 0 2 0 1 0 0 0 0 3 0 0 0 1 0 0 0 0 2 0 0 0 1 0 0 0 0 3 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 9 9 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 3 2 0 0

Sample Output

18 2 17 12 0 52 43

Source

Japan 2006 做的第一道的插頭DP題目。 插頭DP很難理解，看了很久才看懂別人的程序，然后自己寫了下，稍微理解了一點。 決定快速學會插頭DP，寫個插頭DP的總結(jié)，然后學習下概率DP。 應該算是比較基礎的插頭DP的題目了。就是要把兩個2，和兩個3連起來，問經(jīng)過的最少格子數(shù)-2。 普通的插頭DP狀態(tài)轉(zhuǎn)移。 /* POJ 3133 G++ 782ms 1436K */#include<stdio.h> #include<iostream> #include<string.h> #include<algorithm> using namespace std;const int hash_size=60007; const int INF=100000;int n,m; int map[20][20]; int Pow[40];struct Node {int hash_chart[hash_size],sz;int msk[hash_size];int dp[hash_size];int next[hash_size];void clear(){sz=0;memset(hash_chart,-1,sizeof(hash_chart));}inline void push(int _msk,int val){int x=_msk%hash_size;for(int i=hash_chart[x];i!=-1;i=next[i]){if(msk[i]==_msk){if(dp[i]>val)dp[i]=val;return;}}msk[sz]=_msk;dp[sz]=val;next[sz]=hash_chart[x];hash_chart[x]=sz++;}inline int res()//得到答案 {int x=0;for(int i=hash_chart[x];i!=-1;i=next[i])if(!msk[i])return dp[i]-2;return 0;} }hh[2];//兩個循環(huán)轉(zhuǎn)移狀態(tài)void solve() {for(int i=0;i<n;i++)for(int j=0;j<m;j++)scanf("%d",&map[i][j]);int now,pre;pre=0;now=1;hh[pre].clear();hh[pre].push(0,0);for(int i=0;i<n;i++)for(int j=0;j<=m;j++){hh[now].clear();for(int p=0;p<hh[pre].sz;p++)//從pre的所有狀態(tài)推到now的狀態(tài) {int k=hh[pre].msk[p];//3進制表示的當前的插頭狀態(tài)int t=hh[pre].dp[p];if(j==m){if(k/Pow[m])continue;//最后有插頭不能轉(zhuǎn)移hh[now].push(k*3,t);continue;}int t1=(k/Pow[j])%3;//左int t2=(k/Pow[j+1])%3;//上int tk;if(map[i][j]==0)//當前格子為空 {if(t1==0&&t2==0)//沒有插頭 {tk=k+Pow[j]+Pow[j+1];//增加2號插頭hh[now].push(tk,t+1);tk=k+(Pow[j]<<1)+(Pow[j+1]<<1);//增加3號插頭hh[now].push(tk,t+1);tk=k;//不加插頭 hh[now].push(tk,t);}else if((t1&&(!t2))||(t2&&(!t1)))//只有一個插頭 {int temp=k-t1*Pow[j]-t2*Pow[j+1];int temps=(!t1)?t2:t1;tk=temp+temps*Pow[j];//插頭從下邊出來hh[now].push(tk,t+1);tk=temp+temps*Pow[j+1];//插頭從右邊出來hh[now].push(tk,t+1);}else if((t1==t2)&&t1)//有兩個一樣的插頭 {tk=k-t1*Pow[j]-t2*Pow[j+1];//把插頭消去hh[now].push(tk,t+1);}}else if(map[i][j]==1)//障礙 {if(t1==0&&t2==0)//不能有插頭 {tk=k;hh[now].push(tk,t);}}else if(map[i][j]==2)//2號 {if(t1==0&&t2==0){tk=k+Pow[j];hh[now].push(tk,t+1);tk=k+Pow[j+1];hh[now].push(tk,t+1);}else if((t1==1&&t2==0)||(t1==0&&t2==1)){tk=k-Pow[j]*t1-Pow[j+1]*t2;hh[now].push(tk,t+1);}}else if(map[i][j]==3){if(t1==0&&t2==0){tk=k+(Pow[j]<<1);hh[now].push(tk,t+1);tk=k+(Pow[j+1]<<1);hh[now].push(tk,t+1);}else if((t1==2&&t2==0)||(t1==0&&t2==2)){tk=k-Pow[j]*t1-Pow[j+1]*t2;hh[now].push(tk,t+1);}}}swap(now,pre);}printf("%d\n",hh[pre].res()); } int main() {//freopen("in.txt","r",stdin);//freopen("out.txt","w",stdout);Pow[0]=1;for(int i=1;i<20;i++)Pow[i]=3*Pow[i-1];while(scanf("%d%d",&n,&m)){if(n==0&&m==0)break;solve();}return 0; }

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