hdu1964之插頭DP求最優值

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

 

題意有點難理解，直接看輸入好了，空格表示可通過的點,數字表示相鄰點之間通過所需要的花費

問通過所有點並回到原點所需要的最少花費

插頭DP模板題，只需要將到達某個狀態的數量改成最少花費即可

#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的花費 
int head[MAX],next[MAX],Hash[MAX];
LL dp[2][MAX],state[2][MAX],sum;//dp記錄到達相應狀態的最少花費 

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;
	//頭插法
	Hash[size]=total[index];
	next[size]=head[pos];
	head[pos]=size++; 
}

void DP(){//採用4進制括號表示法 
	for(int i=1;i<=n;++i){
		for(int k=1;k<=total[index];++k)state[index][k]<<=2;//由上一行到本行最後一個插頭(0)去掉,在頭增加一個插頭(0) 
		for(int j=1;j<=m;++j){//決策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){//上一格的有效狀態數 
				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;
				//這裏就不用判斷mp[i][j]是否可以通過,因爲這裏一定能通過 
				if(!p && !q){//創建新連通塊
					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;
}