HDU1964 Pipes —— 插头DP

题目链接：https://vjudge.net/problem/HDU-1964

Pipes

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 987    Accepted Submission(s): 494

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

 

Source

NWERC2004

 

Recommend

wangye

题意：

给出一个n*m（n、m<=10）的棋盘，对于一个格子，分别给出它与四个方向相连所需要的花费（即打穿这堵墙所需的花费），求一个回路使得这个回路经过所有的格子刚好一次，且花费是最小的，输出最小花费。

题解：

与此题（URAL1519 Formula 1）无异，只不过把统计个数改成求最小值。

代码如下：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <vector>
 #include <cmath>
 #include <queue>
 #include <stack>
 #include <map>
 #include <string>
 #include <set>
 using namespace std;
 typedef long long LL;
 const int INF = 2e9;
 const LL LNF = 9e18;
 const int MOD = 1e9+;
 const int MAXN = 1e5;
 const int HASH = 1e4;

 int n, m, last_x, last_y;
 bool maze[][];
 int down_cost[][], ri_cost[][];

 struct
 {
     int size, head[HASH], next[MAXN];
     LL state[MAXN];
     int cost[MAXN];

     void init()
     {
         size = ;
         memset(head, -, sizeof(head));
     }

     void insert(LL status, int Cost)
     {
         int u = status%HASH;
         for(int i = head[u]; i!=-; i = next[i])
         {
             if(state[i]==status)
             {
                 cost[i] = min(cost[i], Cost);
                 return;
             }
         }
         state[size] = status;
         cost[size] = Cost;
         next[size] = head[u];
         head[u] = size++;
     }

 }Hash_map[];

 struct
 {
     int code[];
     LL encode(int m)
     {
         LL status = ;
         int id[], cnt = ;
         memset(id, -, sizeof(id));
         id[] = ;
         for(int i = m; i>=; i--)
         {
             if(id[code[i]]==-) id[code[i]] = ++cnt;
             code[i] = id[code[i]];
             status <<= ;
             status += code[i];
         }
         return status;
     }

     void decode(int m, LL status)
     {
         memset(code, , sizeof(code));
         for(int i = ; i<=m; i++)
         {
             code[i] = status&;
             status >>= ;
         }
     }

     void shift(int m)
     {
         for(int i = m-; i>=; i--)
             code[i+] = code[i];
         code[] = ;
     }

 }Line;

 //插头的花费在新建的时候加进去
 void transfer(int i, int j, int cur)
 {
     for(int k = ; k<Hash_map[cur].size; k++)
     {
         LL status = Hash_map[cur].state[k];
         LL Cost = Hash_map[cur].cost[k];
         Line.decode(m, status);
         int up = Line.code[j];
         int left = Line.code[j-];

         if(!up && !left)
         {
             if(maze[i+][j] && maze[i][j+])
             {
                 Line.code[j] = Line.code[j-] = ;  //最多只有5个连通分量
                 Hash_map[cur^].insert(Line.encode(m), Cost+down_cost[i][j]+ri_cost[i][j]);
             }
         }
         else if( (left&&!up) || (!left&&up) )
         {
             int line = left?left:up;
             if(maze[i][j+])
             {
                 Line.code[j-] = ;
                 Line.code[j] = line;
                 Hash_map[cur^].insert(Line.encode(m), Cost+ri_cost[i][j]);
             }
             if(maze[i+][j])
             {
                 Line.code[j-] = line;
                 Line.code[j] = ;
                 if(j==m) Line.shift(m);
                 Hash_map[cur^].insert(Line.encode(m), Cost+down_cost[i][j]);
             }
         }
         else
         {
             if(up!=left)
             {
                 Line.code[j] = Line.code[j-] = ;
                 for(int t = ; t<=m; t++)
                     if(Line.code[t]==up)
                         Line.code[t] = left;
                 if(j==m) Line.shift(m);
                 Hash_map[cur^].insert(Line.encode(m), Cost);
             }
             else if(i==last_x && j==last_y)
             {
                 Line.code[j] = Line.code[j-] = ;
                 if(j==m) Line.shift(m);
                 Hash_map[cur^].insert(Line.encode(m), Cost);
             }
         }
     }
 }

 int main()
 {
     int T;
     char str[];
     scanf("%d", &T);
     while(T--)
     {
         scanf("%d%d", &n, &m); getchar();
         memset(maze, false, sizeof(maze));
         for(int i = ; i<=n; i++)
         for(int j = ; j<=m; j++)
             maze[i][j] = true;

         gets(str);
         for(int i = ; i<n; i++)
         {
             gets(str);
             for(int j = ; j<m; j++)
                 ri_cost[i][j] = str[j*] - '';
             gets(str);
             for(int j = ; j<=m; j++)
                 down_cost[i][j] = str[j*-] - '';
         }
         gets(str);
         for(int j = ; j<m; j++)
             ri_cost[n][j] = str[j*] - '';
         gets(str);
         last_x = n;
         last_y = m;

         int cur = ;
         Hash_map[cur].init();
         Hash_map[cur].insert(, );
         for(int i = ; i<=n; i++)
         for(int j = ; j<=m; j++)
         {
             Hash_map[cur^].init();
             transfer(i, j, cur);
             cur ^= ;
         }

         LL last_status = ;
         LL ans = Hash_map[cur].size?Hash_map[cur].cost[last_status]:;
         printf("%d\n", ans);
     }
 }