hdu 1964（插头DP一条回路）

 

Pipes

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 171    Accepted Submission(s): 75

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

 

分析：此题依旧是插头DP一条回路问题，继续套之前那个模板，状态转移改变一下就可以了，不过要注意，状态转移时有不可到达的状态，要判断一下，我因此超时三次啊，悲剧~~~~‍

代码：

#include<cstdio> #include<cstring> using namespace std; const int mm=10007; struct data { int s[mm],h[mm],p[mm],t; __int64 d[mm]; int hash(int x,__int64 dd) { if(dd>2000)return 0; int i,c=x%mm; for(i=h[c];i>=0;i=p[i]) if(s[i]==x) { if(dd<d[i])d[i]=dd; return i; } s[t]=x,p[t]=h[c],h[c]=t,d[t]=dd; return t++; } void clear() { t=0,memset(h,-1,sizeof(h)); } }f[2]; int i,j,g1,g2,u,k,n,m,a,b,a1,a2,b1,b2,g[15][15][15][15]; inline int eat(int s,int a,int b,int c,bool f) { int n=1,x; while(n) { if(f)a<<=2,b<<=2,c<<=2; else a>>=2,b>>=2,c>>=2; x=s&c; if(x==a)++n; if(x==b)--n; } return (s^b)|a; } void work(int S,__int64 d) { int x,y; x=a&S,y=b&S; if(x==0&&y==0)f[g2].hash(S|a1|b2,d+g[i][j][i][j+1]+g[i][j][i+1][j]); else if(x==0&&y==b1)f[g2].hash(S,d+g[i][j][i][j+1]),f[g2].hash((S^b1)|a1,d+g[i][j][i+1][j]); else if(x==a1&&y==0)f[g2].hash(S,d+g[i][j][i+1][j]),f[g2].hash((S^a1)|b1,d+g[i][j][i][j+1]); else if(x==a1&&y==b1)f[g2].hash(eat(S^a1^b1,b1,b2,b,1),d); else if(x==0&&y==b2)f[g2].hash(S,d+g[i][j][i][j+1]),f[g2].hash((S^b2)|a2,d+g[i][j][i+1][j]); else if(x==a2&&y==0)f[g2].hash(S,d+g[i][j][i+1][j]),f[g2].hash((S^a2)|b2,d+g[i][j][i][j+1]); else if(x==a2&&y==b2)f[g2].hash(eat(S^a2^b2,a2,a1,a,0),d); else if(x==a2&&y==b1)f[g2].hash(S^a2^b1,d); } __int64 DP() { f[0].clear(); f[0].hash(0,0); for(--n,--m,g1=1,g2=i=0;i<=n;++i) { for(k=0;k<f[g2].t;++k)f[g2].s[k]<<=2; a=3,b=3<<2,a1=1,a2=2,b1=1<<2,b2=2<<2; for(j=0;j<=m;a<<=2,b<<=2,a1<<=2,a2<<=2,b1<<=2,b2<<=2,++j) for(g1=!g1,g2=!g2,f[g2].clear(),k=0;k<f[g1].t;++k) { if(i==n&&j==m) { if(f[g1].s[k]==(a1|b2))return f[g1].d[k];else continue; } work(f[g1].s[k],f[g1].d[k]); } } return 0; } inline void get(int & a) { char ch=getchar(); while (ch<'0'||ch>'9')ch=getchar(); for(a=0;ch>='0'&&ch<='9';ch=getchar())a=a*10+ch-48; } int main() { freopen("a.in","r",stdin); int T; get(T); while(T--) { get(n),get(m); memset(g,10,sizeof(g)); for(i=0;i<n;++i) { for(j=1;j<m;++j)get(g[i][j-1][i][j]); if(i<n-1)for(j=0;j<m;++j)get(g[i][j][i+1][j]); } printf("%I64d/n",DP()); } return 0; }