陈老师的多校联合20140811||HDU 3236 ||2009年武汉站G题 01背包问题

http://acm.hdu.edu.cn/showproblem.php?pid=3236

Problem Description

After winning two coupons for the largest shopping mart in your city, you can't wait inviting your girlfriend for gift hunting. Having inspected hundreds of kinds of souvenirs, toys and cosmetics, you finally narrowed down the candidate list to only  n gifts, numbered 1 to  n. Each gift has a happiness value that measures how happy your girlfriend would be, if you get this gift for her. Some of them are special - you  must get it for your girlfriend (note that whether a gift is special has nothing to do with its happiness value).

Coupon 1 can be used to buy gifts with total price not greater than  V1 (RMB). Like most other coupons, you  can’t get any money back if the total price is strictly smaller than  V1. Coupon 2 is almost the same, except that it’s worth  V2. Coupons should be used separately. That means you cannot combine them into a super-coupon that’s worth  V1+ V2. You have to divide the gifts you choose into two part, one uses coupon 1, the other uses coupon 2.

It is your girlfriend's birthday today. According to the rules of the mart, she can take one (only one) gift for FREE! Here comes your challenge: how to make your girlfriend as happy as possible?

 

Input

There will be at most 20 test cases. Each case begins with 3 integers  V1,  V2 and  n (1 <=  V1 <= 500, 1 <=  V2 <= 50, 1 <=  n <= 300), the values of coupon 1 and coupon 2 respectively, and the number of candidate gifts. Each of the following  n lines describes a gift with 3 integers:  P,  H and  S, where  P is the price,  H is the happiness (1 <=  P,H <= 1000),  S=1 if and only if this is a special gift - you must buy it (or get it for free). Otherwise  S=0. The last test case is followed by  V1 =  V2 =  n = 0, which should not be processed.

 

Output

For each test case, print the case number and the maximal total happiness of your girlfriend. If you can't finish the task, i.e. you are not able to buy all special gifts even with the 1-FREE bonus, the happiness is -1 (negative happiness means she's unhappy). Print a blank line after the output of each test case.

 

Sample Input

  
  

   
   3 2 4
3 10 1
2 10 0
5 100 0
5 80 0
3 2 4
3 10 1
2 10 0
5 100 0
5 80 1
0 0 0
  
  

 

Sample Output

  
  

   
   Case 1: 120

Case 2: 100
  
  
  
  


  
  

题目大意：某人又两张优惠券可以购买的面值为v1和v2，一共有n个物品。其中每个物品价格为a，价值为b，属性为0或1，属性为1必须买，此人还有一次可以免费获得一件物品的机会，问在所有必买的物品都买到的前提下获得最大价值是多少？

解题思路：dp[i][j][k][0，1]表示前i物品已经选了，第一张优惠劵使用j元钱，第二张优惠劵使用了k元钱，一次免费的权利是否使用的状态下能得到的最大快乐值。用到滚动数组，空间太大。(代码借鉴)

#include<stdio.h>
#include<iostream>
#include<string.h>
#define inff 0x3fffffff
using namespace std;
int v1,v2,n;
int p[310],h[310],s[310];
int dp[2][510][55][2];//滚动数组优化空间
int main()
{
    int i,j,k;
    int cas=1;
    while(scanf("%d%d%d",&v1,&v2,&n)&&v1+v2+n)
    {
        for(i=1; i<=n; i++)
            scanf("%d%d%d",&p[i],&h[i],&s[i]);
        for(i=0; i<=1; i++) //初始化
            for(j=0; j<=v1; j++)
                for(k=0; k<=v2; k++)
                    dp[i][j][k][0]=dp[i][j][k][1]=-inff;
        dp[0][0][0][0]=0;
        int t=1;
        for(i=1; i<=n; i++)
        {
            for(j=0; j<=v1; j++)
            {
                for(k=0; k<=v2; k++)
                {
                    if(s[i]==0)//不买这个商品
                    {
                        dp[t][j][k][0]=dp[1-t][j][k][0];
                        dp[t][j][k][1]=dp[1-t][j][k][1];
                    }
                    if(j-p[i]>=0)//用第一种优惠劵买
                    {
                        dp[t][j][k][0]=max(dp[t][j][k][0],dp[1-t][j-p[i]][k][0]+h[i]);
                        dp[t][j][k][1]=max(dp[t][j][k][1],dp[1-t][j-p[i]][k][1]+h[i]);
                    }
                    if(k-p[i]>=0)//用第二种优惠劵买
                    {
                        dp[t][j][k][0]=max(dp[t][j][k][0],dp[1-t][j][k-p[i]][0]+h[i]);
                        dp[t][j][k][1]=max(dp[t][j][k][1],dp[1-t][j][k-p[i]][1]+h[i]);
                    }
                    dp[t][j][k][1]=max(dp[t][j][k][1],dp[1-t][j][k][0]+h[i]);//免费获得这个商品
                }
            }
            t=1-t;
            for(j=0; j<=v1; j++)
                for(k=0; k<=v2; k++)
                    dp[t][j][k][0]=dp[t][j][k][1]=-inff;
        }
        int ans=-inff;
        for(i=0; i<=v1; i++)
            for(j=0; j<=v2; j++)
            {
                ans=max(ans,dp[1-t][i][j][0]);
                ans=max(ans,dp[1-t][i][j][1]);
            }
        printf("Case %d: %d\n",cas++,max(-1,ans));
        puts("");
    }
    return 0;
}