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hdu1964之插头DP求最优值

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12/02
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Pipes

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 477    Accepted Submission(s): 238

Problem Description
The construction of office buildings has become a very standardized task. Pre-fabricated modules are combined according to the customer’s needs, shipped from a faraway factory, and assembled on the construction site. However, there are still some tasks that
require careful planning, one example being the routing of pipes for the heating system.



Amodern office building ismade up of squaremodules, one on each floor being a service module from which (among other things) hot water is pumped out to the other modules through the heating pipes. Each module (including the service module) will have heating
pipes connecting it to exactly two of its two to four neighboring modules. Thus, the pipes have to run in a circuit, from the service module, visiting each module exactly once, before finally returning to the service module. Due to different properties of
the modules, having pipes connecting a pair of adjacent modules comes at different costs. For example, some modules are separated by thick walls, increasing the cost of laying pipes. Your task is to, given a description of a floor of an office building, decide
the cheapest way to route the heating pipes.
Input
The first line of input contains a single integer, stating the number of floors to handle. Then follow n floor descriptions, each beginning on a new line with two integers, 2 <= r <= 10 and 2 <= c <= 10, defining the size of the floor – r-by-c modules. Beginning
on the next line follows a floor description in ASCII format, in total 2r + 1 rows, each with 2c + 2 characters, including the final newline. All floors are perfectly rectangular, and will always have an even number of modules. All interior walls are represented
by numeric characters, ’0’ to ’9’, indicating the cost of routing pipes through the wall (see sample input).
Output
For each test case, output a single line with the cost of the cheapest route.
Sample Input 
3

4 3

#######

# 2 3 #

#1#9#1#

# 2 3 #

#1#7#1#

# 5 3 #

#1#9#1#

# 2 3 #

#######

4 4

#########

# 2 3 3 #

#1#9#1#4#

# 2 3 6 #

#1#7#1#5#

# 5 3 1 #

#1#9#1#7#

# 2 3 0 #

#########

2 2

#####

# 1 #

#2#3#

# 4 #

#####
Sample Output 
28

45

10

题意有点难理解，直接看输入好了，空格表示可通过的点,数字表示相邻点之间通过所须要的花费

问通过全部点并回到原点所须要的最少花费

插头DP模板题，仅仅须要将到达某个状态的数量改成最少花费就可以 

#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=100000+10;

const int N=10+10;

int n,m,size,index;

int mp[N][N],w[N][N][N][N],total[2],bit[N];//w[i][j][k][t]表示i,j->k,t的花费

int head[MAX],next[MAX],Hash[MAX];

LL dp[2][MAX],state[2][MAX],sum;//dp记录到达对应状态的最少花费 

void Init(){

	memset(mp,0,sizeof mp);

	for(int i=1;i<=n;++i)for(int j=1;j<=m;++j)mp[i][j]=1;

	index=0;

	sum=INF;

	total[index]=1;

	dp[index][1]=0;

	state[index][1]=0;

}

void HashCalState(LL s,LL v){

	int pos=s%MAX;

	for(int i=head[pos];i != -1;i=next[i]){

		if(state[index][Hash[i]] == s){

			dp[index][Hash[i]]=min(dp[index][Hash[i]],v);

			return;

		}

	}

	++total[index];

	state[index][total[index]]=s;

	dp[index][total[index]]=v;

	//头插法

	Hash[size]=total[index];

	next[size]=head[pos];

	head[pos]=size++;

}

void DP(){//採用4进制括号表示法

	for(int i=1;i<=n;++i){

		for(int k=1;k<=total[index];++k)state[index][k]<<=2;//由上一行到本行最后一个插头(0)去掉,在头添加一个插头(0)

		for(int j=1;j<=m;++j){//决策i,j格

			memset(head,-1,sizeof head);

			size=0;

			index=index^1;

			total[index]=0;

			for(int k=1;k<=total[index^1];++k){//上一格的有效状态数

				LL s=state[index^1][k];

				LL v=dp[index^1][k];

				int p=(s>>bit[j-1])%4;

				int q=(s>>bit[j])%4;

				//这里就不用推断mp[i][j]能否够通过,由于这里一定能通过

				if(!p && !q){//创建新连通块

					if(!mp[i][j+1] || !mp[i+1][j])continue;

					s=s+(1<<bit[j-1])+2*(1<<bit[j]);

					v=v+w[i][j][i][j+1]+w[i][j][i+1][j];

					HashCalState(s,v);

				}else if(!p && q){

					if(mp[i][j+1])HashCalState(s,v+w[i][j][i][j+1]);

					if(mp[i+1][j]){

						s=s+q*(1<<bit[j-1])-q*(1<<bit[j]);

						v=v+w[i][j][i+1][j];

						HashCalState(s,v);

					}

				}else if(p && !q){

					if(mp[i+1][j])HashCalState(s,v+w[i][j][i+1][j]);

					if(mp[i][j+1]){

						s=s-p*(1<<bit[j-1])+p*(1<<bit[j]);

						v=v+w[i][j][i][j+1];

						HashCalState(s,v);

					}

				}else if(p == 1 && q == 1){

					int b=1;

					for(int t=j+1;t<=m;++t){

						int a=(s>>bit[t])%4;

						if(a == 1)++b;

						if(a == 2)--b;

						if(!b){

							s=s+(1<<bit[t])-2*(1<<bit[t]);

							break;

						}

					}

					s=s-(1<<bit[j-1])-(1<<bit[j]);

					HashCalState(s,v);

				}else if(p == 2 && q == 2){

					int b=1;

					for(int t=j-2;t>=0;--t){

						int a=(s>>bit[t])%4;

						if(a == 2)++b;

						if(a == 1)--b;

						if(!b){

							s=s-(1<<bit[t])+2*(1<<bit[t]);

							break;

						}

					}

					s=s-2*(1<<bit[j-1])-2*(1<<bit[j]);

					HashCalState(s,v);

				}else if(p == 1 && q == 2){

					if(i == n && j == m)sum=min(sum,v);

				}else if(p == 2 && q == 1){

					s=s-2*(1<<bit[j-1])-(1<<bit[j]);

					HashCalState(s,v);

				}

			}

		}

	}

}

int main(){

	for(int i=0;i<N;++i)bit[i]=i<<1;

	int t;

	char s[N+N];

	scanf("%d",&t);

	while(t--){

		scanf("%d%d%*c",&n,&m);

		Init();

		gets(s);

		for(int i=1;i<=n;++i){

			gets(s);

			for(int j=2;j<2*m+1;j+=2){

				w[i][j/2][i][j/2+1]=s[j]-'0';

			}

			gets(s);

			for(int j=1;j<2*m+1;j+=2){

				w[i][j/2+1][i+1][j/2+1]=s[j]-'0';

			}

		}

		DP();

		printf("%lld\n",sum);

	}

	return 0;

}

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