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
机智的我虽然看出是这个分类依旧卡了3个多点的题，状态好次==
题意：给女朋友买一堆礼物，每个礼物有一个快乐值，有n个备选的，其中有一部分是必买的，总共有两张支票，价值v1,v2，花钱少了不给退，两张支票不能凑在一起用，而且有一个礼物是免费的，问女友最多可以多快乐？
做法：最开始的思路是：dp[v1][v2]，三重循环，一维是礼物，二维、三维分别是dp的两维，然后循环的时候犯了一个低级错误，j,k正常来说是大于cost[i]循环到cost[i[就停止了，但是这个题不能停啊！比方说j循环到cost[i[停了，但是k不一定不满足啊，这个题是两维啊！
然后，题中要求有一个是免费的，我受之前做过的题（貌似是饭卡吧）的影响，纠结于是要找出一个最贵的还是最快乐的？觉得是应该找出来一个“快乐值”最大的，然而并不是的。这个题的处理方法是再加一维dp表示是否用过这个免费的机会（貌似之前树形dp做过吧）之后，dp转移是很容易可以想到的==
#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int cost1[309],hap1[309],cost0[309],hap0[309],dp[505][55][2],n,v1,v2;
int main()
{
// freopen("cin.txt","r",stdin);
//freopen("out.txt","w",stdout);
int cas=1;
while(~scanf("%d%d%d",&v1,&v2,&n))
{
if(v1==0&&v2==0&&n==0)break;
memset(dp,0,sizeof(dp));
int l0=0,l1=0,maxh=0,maxn;
for(int i=0;i<n;i++)
{
int a,b,c;
scanf("%d%d%d",&a,&b,&c);
if(c==1)
{
cost1[l1]=a;
hap1[l1++]=b;
maxh+=b;
}
if(c==0)
{
cost0[l0]=a;
hap0[l0++]=b;
}
}
for(int i=0;i<l1;i++)
{
for(int j=v1;j>=0;j--)
{
for(int k=v2;k>=0;k--)
{
dp[j][k][1]=max(dp[j][k][1],dp[j][k][0]+hap1[i]);
if(j>=cost1[i])
dp[j][k][0]=max(dp[j][k][0],dp[j-cost1[i]][k][0]+hap1[i]),
dp[j][k][1]=max(dp[j][k][1],dp[j-cost1[i]][k][1]+hap1[i]);
if(k>=cost1[i])
dp[j][k][0]=max(dp[j][k][0],dp[j][k-cost1[i]][0]+hap1[i]),
dp[j][k][1]=max(dp[j][k][1],dp[j][k-cost1[i]][1]+hap1[i]);
// printf("i=%d,j=%d,k=%d,dp0=%d,dp1=%d\n",i,j,k,dp[j][k][0],dp[j][k][1]);
}
}
}
if(dp[v1][v2][1]<maxh)maxn=-1;
else
{
for(int i=0;i<l0;i++)
{
for(int j=v1;j>=0;j--)
{
for(int k=v2;k>=0;k--)
{
if(dp[j][k][0]>=maxh)
dp[j][k][1]=max(dp[j][k][1],dp[j][k][0]+hap0[i]);
if(j>=cost0[i]&&dp[j-cost0[i]][k][0]>=maxh)
dp[j][k][0]=max(dp[j][k][0],dp[j-cost0[i]][k][0]+hap0[i]);
if(j>=cost0[i]&&dp[j-cost0[i]][k][1]>=maxh)
dp[j][k][1]=max(dp[j][k][1],dp[j-cost0[i]][k][1]+hap0[i]);
if(k>=cost0[i]&&dp[j][k-cost0[i]][0]>=maxh)
dp[j][k][0]=max(dp[j][k][0],dp[j][k-cost0[i]][0]+hap0[i]);
if(k>=cost0[i]&&dp[j][k-cost0[i]][1]>=maxh)
dp[j][k][1]=max(dp[j][k][1],dp[j][k-cost0[i]][1]+hap0[i]);
// printf("i=%d,j=%d,k=%d,dp0=%d,dp1=%d\n",i,j,k,dp[j][k][0],dp[j][k][1]);
}
}
}
maxn=dp[v1][v2][1];
}
printf("Case %d: %d\n\n",cas++,maxn);
}
return 0;
}