HDu 3466 Proud Merchants 01リュックサック変形
7362 ワード
Proud Merchants
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others) Total Submission(s): 1978 Accepted Submission(s): 792
Problem Description
Recently, iSea went to an ancient country. For such a long time, it was the most wealthy and powerful kingdom in the world. As a result, the people in this country are still very proud even if their nation hasn’t been so wealthy any more. The merchants were the most typical, each of them only sold exactly one item, the price was Pi, but they would refuse to make a trade with you if your money were less than Qi, and iSea evaluated every item a value Vi. If he had M units of money, what’s the maximum value iSea could get?
Input
There are several test cases in the input.
Each test case begin with two integers N, M (1 ≤ N ≤ 500, 1 ≤ M ≤ 5000), indicating the items’ number and the initial money. Then N lines follow, each line contains three numbers Pi, Qi and Vi (1 ≤ Pi ≤ Qi ≤ 100, 1 ≤ Vi ≤ 1000), their meaning is in the description.
The input terminates by end of file marker.
Output
For each test case, output one integer, indicating maximum value iSea could get.
Sample Input
2 10
10 15 10
5 10 5
3 10
5 10 5
3 5 6
2 7 3
Sample Output
5
11
Author
iSea @ WHU
Source
2010 ACM-ICPC Multi-University Training Contest（3）——Host by WHU
标题:買い物、pi qi vi △piは使うお金を表し、qiはあなたの現在のお金がこの値以上でなければ買えないことを表しています.viの価値です.
単純な01リュックと比較するとqiが多くなり、ここで違いが発生します.
for(j=sum_money;j>=f[i].qi&&(j-f[i].vi)>=0;j--)
qi−piは、差分値が小さいほど先に置かれるように、小さいから大きいまで並べ替えられる.
どうしてこのように並べ替えますか?手を出して描きなさい.詳しい証明は後で書きます.
実は単純な01リュックの差は0なので、共通性があります.
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others) Total Submission(s): 1978 Accepted Submission(s): 792
Problem Description
Recently, iSea went to an ancient country. For such a long time, it was the most wealthy and powerful kingdom in the world. As a result, the people in this country are still very proud even if their nation hasn’t been so wealthy any more. The merchants were the most typical, each of them only sold exactly one item, the price was Pi, but they would refuse to make a trade with you if your money were less than Qi, and iSea evaluated every item a value Vi. If he had M units of money, what’s the maximum value iSea could get?
Input
There are several test cases in the input.
Each test case begin with two integers N, M (1 ≤ N ≤ 500, 1 ≤ M ≤ 5000), indicating the items’ number and the initial money. Then N lines follow, each line contains three numbers Pi, Qi and Vi (1 ≤ Pi ≤ Qi ≤ 100, 1 ≤ Vi ≤ 1000), their meaning is in the description.
The input terminates by end of file marker.
Output
For each test case, output one integer, indicating maximum value iSea could get.
Sample Input
2 10
10 15 10
5 10 5
3 10
5 10 5
3 5 6
2 7 3
Sample Output
5
11
Author
iSea @ WHU
Source
2010 ACM-ICPC Multi-University Training Contest（3）——Host by WHU
标题:買い物、pi qi vi △piは使うお金を表し、qiはあなたの現在のお金がこの値以上でなければ買えないことを表しています.viの価値です.
単純な01リュックと比較するとqiが多くなり、ここで違いが発生します.
for(j=sum_money;j>=f[i].qi&&(j-f[i].vi)>=0;j--)
qi−piは、差分値が小さいほど先に置かれるように、小さいから大きいまで並べ替えられる.
どうしてこのように並べ替えますか?手を出して描きなさい.詳しい証明は後で書きます.
実は単純な01リュックの差は0なので、共通性があります.
1 #include<cstdio>
2 #include<cstring>
3 #include<cstdlib>
4 #include<algorithm>
5 using namespace std;
6
7 struct node
8 {
9 int pi,qi,vi;
10 }f[502];
11 int dp[5003];
12
13 bool cmp(node n1,node n2)
14 {
15 return (n1.qi-n1.pi)<(n2.qi-n2.pi);
16 }
17
18 int main()
19 {
20 int n,sum_money,max;
21 int i,j,tmp;
22 while(scanf("%d%d",&n,&sum_money)>0)
23 {
24 for(i=1;i<=n;i++)
25 {
26 scanf("%d%d%d",&f[i].pi,&f[i].qi,&f[i].vi);
27 }
28 sort(f+1,f+1+n,cmp);
29 memset(dp,0,sizeof(dp));
30
31 for(i=1;i<=n;i++)
32 {
33 for(j=sum_money;j>=f[i].qi&&(j-f[i].pi)>=0;j--)
34 {
35 tmp=dp[j-f[i].pi]+f[i].vi;
36 if(tmp>dp[j])
37 dp[j]=tmp;
38 }
39 }
40 printf("%d
",dp[sum_money]);
41 }
42 return 0;
43 }