Chemical Reaction - UVa 10604 dp

Chemical Reaction

In a chemist’s lab, there are several types of chemicals in tubes. The chemist wants to mix all these chemicals together, two chemicals at a time. Whenever two chemicals are mixed, some heat is generated and released into the air and the mixed chemical is
a known chemical of possibly other type than the original two. The resulting chemical type and the amount of heats emitted can looked up in the chemical mixture table.

Chemicals	1	2	3
Resulting chemical type	Heats emmited	Resulting chemical type	Heats emmited	Resulting chemical type	Heats emmited
1	1	0	3	-10	3	3000
2	3	-10	2	0	1	-500
3	3	3000	1	-500	3	0

For example, in the above chemical mixture table, there are three types of chemicals: 1, 2, and 3. If you mix chemicals 1 and 3, they produce +3000 units of heat and turn into chemical 3. Sometimes, the heat generated can be negative. For instance, you can
mix 2 and 3 and they turn into chemical 1 and in the meantime, cool down the lab by 500 units of heat. Since the chemist lacks funding to buy necessary equipments to protect himself from the heat generated, it is utmost important to find a way to mix all the
chemicals together that produces the least total heat, regardless of the final chemical type. For example, suppose the lab has four tubes containing chemicals of types 1, 2, 2, and 3. If the chemicals are mixed in the parenthesize order of ((1 2) (2 3)), it
will produce (–10)+ (-500)+(3000) = 2490 units of heat. However, if the chemicals are mixed in the (2 (1 (2 3))) order, it will produce (-500)+0+(-10)= -510 units of heat, which is also the least total heat possible.

Input

The first line of input file consists of a single number denoting the number of test cases in the file. There is a single line containing a ‘/’ character separating two consecutive test cases. The end of the file is marked with a line containing a ‘.’ For
each test case, the first line contains an integer number m (1≤m≤6) denoting the number chemical types. Therefore, the chemicals are indexed from 1 up to at most 6. The next mxm lines give the chemical mixture table entries in the row-major order. Each line
contains the new resulting chemical type and the amount of energy emitted. After the table entries is a line with a single number k (2≤k≤10) denoting number of tubes in the lab. The next line then contains k integers in the range of 1 and m, separated by blank
spaces, denoting the types of chemicals in those k tubes.

Output

For each test case, output the least total heat possible on a single line.

Sample Input	Sample Output
2 3 1 0 3 -10 3 3000 3 -10 2 0 1 -500 3 3000 1 -500 3 0 4 1 2 2 3 / 3 1 0 3 500 3 -250 3 500 2 0 1 100 3 -250 1 100 3 0 6 1 1 1 2 2 3 .	-510 -650

题意：混合任意两瓶药水，会产生一瓶新的药水，然后会放出或吸收一定的能量，问最后合并到只剩一瓶时最少放出多少能量。

思路：看了下其他人的代码，基本上都是六维的dp……还好我有独立思考的习惯……于是乎……这道题被我做成了11进制的状压dp %>_<%，dp[S]表示在S状态下最少放出的药水，vector <int> vc[i]表示在剩i瓶药水时有哪些情况。注意题目有坑，就是a类型和b类型混合，与b类型和a类型混合可能不一样（真坑，后来队友就给我讲浓硫酸和水的故事……然后就感觉自己二了）。时间挺快，0.029秒，好不容易又有排名靠前的题了。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
using namespace std;
int trans[15][15][2],dp[20000100],vis[20000100],pow11[15],T,t;
vector <int> vc[15];
char s[110];
void solve(int S,int m1,int m2,int num)
{ int type,eng,S2;
  type=trans[m1][m2][0];
  eng=trans[m1][m2][1];
  S2=S-pow11[m1]-pow11[m2]+pow11[type];
  if(vis[S2]==t)
   dp[S2]=min(dp[S2],dp[S]+eng);
  else
  { dp[S2]=dp[S]+eng;
    vis[S2]=t;
    vc[num-1].push_back(S2);
  }
}
int main()
{ int n,m,i,j,k,m1,m2,ans,S,len;
  pow11[1]=1;
  for(i=2;i<=7;i++)
   pow11[i]=pow11[i-1]*11;
  scanf("%d",&T);
  for(t=1;t<=T;t++)
  { scanf("%d",&m);
    for(i=1;i<=m;i++)
     for(j=1;j<=m;j++)
      scanf("%d%d",&trans[i][j][0],&trans[i][j][1]);
    scanf("%d",&n);
    for(i=0;i<=n;i++)
     vc[i].clear();
    S=0;
    for(i=1;i<=n;i++)
    { scanf("%d",&k);
      S+=pow11[k];
    }
    scanf("%s",s);
    dp[S]=0;
    vc
.push_back(S);
    for(i=n;i>=2;i--)
    { len=vc[i].size();
      for(j=0;j<len;j++)
      { S=vc[i][j];
        for(m1=1;m1<=m;m1++)
        { if((S%pow11[m1+1])/pow11[m1]>=2)
           solve(S,m1,m1,i);
          for(m2=m1+1;m2<=m;m2++)
           if((S%pow11[m1+1])/pow11[m1]>0 && (S%pow11[m2+1])/pow11[m2]>0)
           { solve(S,m1,m2,i);
             solve(S,m2,m1,i);
           }
        }
      }
    }
    ans=1000000000;
    len=vc[1].size();
    for(i=0;i<vc[1].size();i++)
     ans=min(ans,dp[vc[1][i]]);
    printf("%d\n",ans);
  }
}