TZOJ 2289 Help Bob(状压DP)

描述

Bob loves Pizza but is always out of money. One day he reads in the newspapers that his favorite pizza restaurant, Alfredo's Pizza Restaurant, is running a competition: they will donate a big pizza to the first person who will tell them the lowest price per area that can be achieved by buying any of the pizzas at most once. "That task is easy!", thinks Bob, "For each pizza I just calculate the average price and the lowest quotient will be the answer.".

Unfortunately the problem is a bit more complicated: with some pizzas Alberto gives out discount coupons for getting another pizza cheaper and even worse, those coupons can be combined. The pizzas have to be bought one after the other, and it is not possible to use a coupon to get a discount retrospectively for a pizza which has already been bought. Can you help Bob to become the first to solve this task and to get a pizza for free?

输入

The input file contains several test cases. Each test case starts with a number m, the number of pizzas Alfredo offers. Input is terminated by m = 0. Otherwise, 1 ≤ m ≤ 15. Then follow m lines describing the pizzas. Each of those following lines describes pizza i (1 ≤ i ≤ m) and starts with 3 integer numbers p_i, a_i and n_i specifying the price of the pizza, its area and the number of discount coupons you get when buying it, 1 ≤ p_i ≤ 10000, 1 ≤ a_i ≤ 10000 and 0 ≤ n_i < m. Then follow n_i pairs of integer numbers x_i,j and y_i,j specifying the index x_i,j (1 ≤ x_i,j ≤ m, x_i,j ≠ i) of the pizza you get a discount coupon for and the discount in percentage terms y_i,j (1 ≤ y_i,j ≤ 50) you get when buying pizza x_i,j. You may assume that for each i the values x_i,j are pairwise distinct.

输出

For each test case print one line containing the lowest price per area that can be achieved by buying any of the pizzas at most once. Round this number to 4 places after the decimal point. Note that you can combine an arbitrary number of discount coupons: for a pizza with price 10 and two rabatt coupons for that pizza with a 50 and a 20 on it, you would only have to pay 10 * 0.8 * 0.5 = 4 monetary units.

样例输入

1
80 30 0
2
200 100 1 2 50
200 100 0
5
100 100 2 3 50 2 50
100 100 1 4 50
100 100 1 2 40
600 600 1 5 10
1000 10 1 1 50
0

样例输出

2.6667
1.5000
0.5333

题意

N个披萨，p价格a面积k送的优惠券数，接着k个代表买x披萨打10-y/10折

每个披萨最多买1个，问怎么买总价格/总面积最大，并输出值

题解

状压DP

dp[i]代表买了i中二进制为1的披萨花的最少钱

那么答案就是min(dp[i]/状态i中二进制为1的披萨总面积)

代码

 #include<bits/stdc++.h>
 using namespace std;

 const int N=(<<);
 int main()
 {
     int n,p[],a[],k[],x[][];
     double dp[N],y[][];
     while(scanf("%d",&n)!=EOF,n)
     {
         for(int i=;i<<<n;i++)dp[i]=1e18;
         for(int i=;i<n;i++)
         {
             scanf("%d%d%d",&p[i],&a[i],&k[i]);
             for(int j=;j<k[i];j++)
             {
                 scanf("%d%lf",&x[i][j],&y[i][j]);
                 x[i][j]--;
                 y[i][j]=-y[i][j]/;
             }
         }
         int state=<<n;
         double minn=1e18;
         dp[]=;
         for(int i=;i<state;i++)
         {
             double co[],area=;
             for(int j=;j<n;j++)co[j]=.;
             for(int j=;j<n;j++)
                 if(i&(<<j))
                 {
                     area+=a[j];
                     for(int l=;l<k[j];l++)
                         co[x[j][l]]*=y[j][l];
                 }
             for(int j=;j<n;j++)
             {
                 if(i&(<<j))continue;
                 dp[i|(<<j)]=min(dp[i|(<<j)],dp[i]+p[j]*co[j]);
                 minn=min(minn,dp[i|(<<j)]/(area+a[j]));
             }
         }
         printf("%.4f\n",minn);
     }
     return ;
 }