poj3133之经典插头DP

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2024-07-02 12:38:14 231人阅读

Manhattan Wiring

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 1482		Accepted: 869

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
…
row_n

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

题意：求连接两个2和两个3的路径之和最小，输出和-2，不存在就输出0

输入0表示可以走，1表示不可以走

分析：按照插头DP的模式进行DP,碰到0时如果p=q=0可以选择不走

碰到2和3的时候保证只有一个插头即可

这题有很多细节要注意

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <queue>
#include <algorithm>
#include <map>
#include <cmath>
#include <iomanip>
#define INF 99999999
typedef long long LL;
using namespace std;

const int MAX=30000+10;
const int maxn=100000+10;
const int N=10+10;
int n,m,size[2],index,bit[N];
int head[MAX],next[maxn];
LL dp[2][maxn],state[2][maxn],sum;
int mp[N][N];

void HashCalState(LL s,LL num){
	int pos=s%MAX;
	for(int i=head[pos];i != -1;i=next[i]){
		if(state[index][i] == s){
			dp[index][i]=min(dp[index][i],num);
			return;
		}
	}
	state[index][size[index]]=s;
	dp[index][size[index]]=num;
	next[size[index]]=head[pos];
	head[pos]=size[index]++; 
}

void DP(){
	//初始化 
	sum=INF;
	index=0;
	size[index]=1;
	state[index][0]=0;
	dp[index][0]=0;
	//逐格进行DP
	for(int i=1;i<=n;++i){
		for(int k=0;k<size[index];++k)state[index][k]<<=2;//上一行最后的空插头减去,本行最前面增加一个空插头,所以*4
		for(int j=1;j<=m;++j){
			memset(head,-1,sizeof head);
			index=index^1;
			size[index]=0;
			for(int k=0;k<size[index^1];++k){
				LL s=state[index^1][k],temp; 
				LL num=dp[index^1][k];
				int p=(s>>bit[j-1])%4;
				int q=(s>>bit[j])%4;
				int w=(s>>bit[j+1])%4;
				if(mp[i][j] == 1){//该格不能走 
					if(!p && !q)HashCalState(s,num);
					continue;
				}else if(!p && !q){
					if(!mp[i][j]){//该点是0需要两个插头或者不走
					 	HashCalState(s,num);//该点是0可以不走
						if(mp[i][j+1] == 1 || mp[i+1][j] == 1)continue;//右边或下边相邻的格子不能到达,无法完成两个插头
						if(mp[i][j+1]+mp[i+1][j] == 5)continue;//表示添加的两个插头两端将是2和3,不能到达 
						if(mp[i][j+1] == 2 || mp[i+1][j] == 2){//有点是2则必须连得插头是1(表示连2的线) 
							temp=s+(1<<bit[j-1])+(1<<bit[j]);
							if(w == 0)HashCalState(temp,num+3);//mp[i][j+1]没有插头则经过该点路径长度+1,即num+2+1 
							else if(w == 1 && mp[i][j+1] != 2)HashCalState(temp,num+2);//增加的路径只有[i+1,j]和[i,j] 
						}
						else if(mp[i][j+1] == 3 || mp[i+1][j] == 3){//有点是3则必须连得插头是2(表示连3的线) 
							temp=s+2*(1<<bit[j-1])+2*(1<<bit[j]);
							if(w == 0)HashCalState(temp,num+3);
							else if(w == 2 && mp[i][j+1] != 3)HashCalState(temp,num+2);//特别要注意w != 0时要判断[i,j+1]是否是2或3,防止2/3的情况下有两个插头 
						}
						else {
							if(w == 0){
								HashCalState(s+(1<<bit[j-1])+(1<<bit[j]),num+3);
								HashCalState(s+2*(1<<bit[j-1])+2*(1<<bit[j]),num+3);
							}else if(w == 1 && mp[i][j+1] != 2)HashCalState(s+(1<<bit[j-1])+(1<<bit[j]),num+2);
							else if(w == 2 && mp[i][j+1] != 3)HashCalState(s+2*(1<<bit[j-1])+2*(1<<bit[j]),num+2);
						}
					}
					else if(mp[i][j] == 2){//只能有独立插头 
						if(mp[i][j+1] != 1 && mp[i][j+1] != 3){
							if(w == 0)HashCalState(s+(1<<bit[j]),num+2);
							else if(w == 1 && mp[i][j+1] != 2)HashCalState(s+(1<<bit[j]),num+1);
						}
						if(mp[i+1][j] != 1 && mp[i+1][j] != 3)HashCalState(s+(1<<bit[j-1]),num+2);
					}
					else if(mp[i][j] == 3){//只能有独立插头 
						if(mp[i][j+1] != 1 && mp[i][j+1] != 2){
							if(w == 0)HashCalState(s+2*(1<<bit[j]),num+2);
							else if(w == 2 && mp[i][j+1] != 3)HashCalState(s+2*(1<<bit[j]),num+1);
						} 
						if(mp[i+1][j] != 1 && mp[i+1][j] != 2)HashCalState(s+2*(1<<bit[j-1]),num+2);
					}
				}else if(!p && q){
					if(mp[i][j]){
						//if(mp[i][j] != q+1)continue;
						s=s-q*(1<<bit[j]);
						HashCalState(s,num);
					}else{
						if(mp[i][j+1] == 0 || mp[i][j+1] == q+1){
							if(w == 0)HashCalState(s,num+1);
							else if(w == q && mp[i][j+1] == 0)HashCalState(s,num);
						}
						if(mp[i+1][j] == 0 || mp[i+1][j] == q+1){
							s=s+q*(1<<bit[j-1])-q*(1<<bit[j]);
							HashCalState(s,num+1);
						}
					}
				}else if(p && !q){
					if(mp[i][j]){
						//if(mp[i][j] != p+1)continue;
						s=s-p*(1<<bit[j-1]);
						HashCalState(s,num);
					}else{
						if(mp[i+1][j] == 0 || mp[i+1][j] == p+1)HashCalState(s,num+1);
						if(mp[i][j+1] == 0 || mp[i][j+1] == p+1){
							s=s-p*(1<<bit[j-1])+p*(1<<bit[j]);
							if(w == 0)HashCalState(s,num+1);
							else if(w == p && mp[i][j+1] == 0)HashCalState(s,num);
						}
					}
				}else if(p == q/*&& !mp[i][j]*/){//p == q == 1或者p == q == 2时p‘,q‘不能有插头 
					s=s-p*(1<<bit[j-1])-q*(1<<bit[j]);
					HashCalState(s,num);
				}
			}
		} 
	}
	//cout<<size[index]<<endl;
	for(int k=0;k<size[index];++k)sum=min(sum,dp[index][k]);
}

int main(){
	for(int i=0;i<N;++i)bit[i]=i<<1;
	while(~scanf("%d%d",&n,&m),n+m){
		for(int i=0;i<N;++i)for(int j=0;j<N;++j)mp[i][j]=1;
		for(int i=1;i<=n;++i){
			for(int j=1;j<=m;++j)cin>>mp[i][j];
		}
		DP();//插头DP
		if(sum == INF)sum=2;
		printf("%lld\n",sum-2);
	}
	return 0;
}
/*
5 9
0 0 0 0 0 0 0 0
0 0 0 3 0 0 0 0
2 0 0 0 0 0 2 0
0 0 0 3 0 0 0 0
0 0 0 0 0 0 0 0

5 6
0 0 0 0 0 0
0 0 0 3 0 0
0 2 0 0 0 2
0 0 0 3 0 0
0 0 0 0 0 0
*/

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poj3133之经典插头DP

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