HDOJ 5410 CRB and His Birthday DP背包
2015-08-21 14:09
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先跑一遍01背包,再跑一遍多重背包
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 327 Accepted Submission(s): 177
Problem Description
Today is CRB's birthday. His mom decided to buy many presents for her lovely son.
She went to the nearest shop with M Won(currency
unit).
At the shop, there are N kinds
of presents.
It costs Wi Won
to buy one present of i-th
kind. (So it costs k × Wi Won
to buy k of
them.)
But as the counter of the shop is her friend, the counter will give Ai × x + Bi candies
if she buys x(x>0)
presents of i-th
kind.
She wants to receive maximum candies. Your task is to help her.
1 ≤ T ≤
20
1 ≤ M ≤
2000
1 ≤ N ≤
1000
0 ≤ Ai, Bi ≤
2000
1 ≤ Wi ≤
2000
Input
There are multiple test cases. The first line of input contains an integer T,
indicating the number of test cases. For each test case:
The first line contains two integers M and N.
Then N lines
follow, i-th
line contains three space separated integers Wi, Ai and Bi.
Output
For each test case, output the maximum candies she can gain.
Sample Input
1
100 2
10 2 1
20 1 1
Sample Output
21
HintCRB's mom buys 10 presents of first kind, and receives 2 × 10 + 1 = 21 candies.
Author
KUT(DPRK)
Source
2015 Multi-University Training Contest 10
/* ***********************************************
Author :CKboss
Created Time :2015年08月21日 星期五 13时29分32秒
File Name :HDOJ5410.cpp
************************************************ */
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>
using namespace std;
const int maxn=1100;
int m,n;
int w[maxn],A[maxn],B[maxn];
int dp[maxn*2];
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int T_T;
scanf("%d",&T_T);
while(T_T--)
{
scanf("%d%d",&m,&n);
for(int i=1;i<=n;i++) scanf("%d%d%d",w+i,A+i,B+i);
memset(dp,0,sizeof(dp));
for(int i=1;i<=n;i++)
{
for(int j=m;j>=w[i];j--)
{
dp[j]=max(dp[j],dp[j-w[i]]+A[i]+B[i]);
}
}
for(int i=1;i<=n;i++)
{
for(int j=w[i];j<=m;j++)
{
dp[j]=max(dp[j],dp[j-w[i]]+A[i]);
}
}
printf("%d\n",dp[m]);
}
return 0;
}
CRB and His Birthday
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 327 Accepted Submission(s): 177
Problem Description
Today is CRB's birthday. His mom decided to buy many presents for her lovely son.
She went to the nearest shop with M Won(currency
unit).
At the shop, there are N kinds
of presents.
It costs Wi Won
to buy one present of i-th
kind. (So it costs k × Wi Won
to buy k of
them.)
But as the counter of the shop is her friend, the counter will give Ai × x + Bi candies
if she buys x(x>0)
presents of i-th
kind.
She wants to receive maximum candies. Your task is to help her.
1 ≤ T ≤
20
1 ≤ M ≤
2000
1 ≤ N ≤
1000
0 ≤ Ai, Bi ≤
2000
1 ≤ Wi ≤
2000
Input
There are multiple test cases. The first line of input contains an integer T,
indicating the number of test cases. For each test case:
The first line contains two integers M and N.
Then N lines
follow, i-th
line contains three space separated integers Wi, Ai and Bi.
Output
For each test case, output the maximum candies she can gain.
Sample Input
1
100 2
10 2 1
20 1 1
Sample Output
21
HintCRB's mom buys 10 presents of first kind, and receives 2 × 10 + 1 = 21 candies.
Author
KUT(DPRK)
Source
2015 Multi-University Training Contest 10
/* ***********************************************
Author :CKboss
Created Time :2015年08月21日 星期五 13时29分32秒
File Name :HDOJ5410.cpp
************************************************ */
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>
using namespace std;
const int maxn=1100;
int m,n;
int w[maxn],A[maxn],B[maxn];
int dp[maxn*2];
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int T_T;
scanf("%d",&T_T);
while(T_T--)
{
scanf("%d%d",&m,&n);
for(int i=1;i<=n;i++) scanf("%d%d%d",w+i,A+i,B+i);
memset(dp,0,sizeof(dp));
for(int i=1;i<=n;i++)
{
for(int j=m;j>=w[i];j--)
{
dp[j]=max(dp[j],dp[j-w[i]]+A[i]+B[i]);
}
}
for(int i=1;i<=n;i++)
{
for(int j=w[i];j<=m;j++)
{
dp[j]=max(dp[j],dp[j-w[i]]+A[i]);
}
}
printf("%d\n",dp[m]);
}
return 0;
}
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