2015暑假多校联合---CRB and His Birthday(01背包)
2016-09-02 20:46
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题目链接
http://acm.split.hdu.edu.cn/showproblem.php?pid=5410
[align=left]Problem Description[/align]
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
[align=left]Input[/align]
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.
[align=left]Output[/align]
For each test case, output the maximum candies she can gain.
[align=left]Sample Input[/align]
1
100 2
10 2 1
20 1 1
[align=left]Sample Output[/align]
21
Hint
CRB's mom buys 10 presents of first kind, and receives 2 × 10 + 1 = 21 candies.
[align=left]Author[/align]
KUT(DPRK)
[align=left]Source[/align]
2015 Multi-University Training Contest 10
[align=left]Recommend[/align]
wange2014
题意:输入M,N,分别表示总的钱数和物品种数,接下来输入N行,每行3个数,单价、买一件送的糖数 、买一次送的糖数 求最多能得到多少糖?
思路:01背包,dp[i]表示i钱下能得到最多的糖数,vis[i][j]表示i钱下得到最多糖时,是否买j物品,状态转移方程dp[i]=dp[i-kind[j][0]]+kind[j][1]+(vis[i-kind[j][0]][j]==0) * kind[j][2];
代码如下:
http://acm.split.hdu.edu.cn/showproblem.php?pid=5410
[align=left]Problem Description[/align]
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
[align=left]Input[/align]
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.
[align=left]Output[/align]
For each test case, output the maximum candies she can gain.
[align=left]Sample Input[/align]
1
100 2
10 2 1
20 1 1
[align=left]Sample Output[/align]
21
Hint
CRB's mom buys 10 presents of first kind, and receives 2 × 10 + 1 = 21 candies.
[align=left]Author[/align]
KUT(DPRK)
[align=left]Source[/align]
2015 Multi-University Training Contest 10
[align=left]Recommend[/align]
wange2014
题意:输入M,N,分别表示总的钱数和物品种数,接下来输入N行,每行3个数,单价、买一件送的糖数 、买一次送的糖数 求最多能得到多少糖?
思路:01背包,dp[i]表示i钱下能得到最多的糖数,vis[i][j]表示i钱下得到最多糖时,是否买j物品,状态转移方程dp[i]=dp[i-kind[j][0]]+kind[j][1]+(vis[i-kind[j][0]][j]==0) * kind[j][2];
代码如下:
#include <iostream> #include <algorithm> #include <cstdio> #include <cstring> #include <cmath> #define eps 1e-8 #define maxn 105 #define inf 0x3f3f3f3f3f3f3f3f #define IN freopen("in.txt","r",stdin); using namespace std; int dp[2005]; int vis[2005][1005]; int kind[1005][3]; int main() { int T; int M,N; cin>>T; while(T--) { scanf("%d%d",&M,&N); for(int i=0;i<N;i++) scanf("%d%d%d",&kind[i][0],&kind[i][1],&kind[i][2]); memset(dp,0,sizeof(dp)); memset(vis,0,sizeof(vis)); for(int i=1;i<=M;i++) { int flag=-1; for(int j=0;j<N;j++) { if(i<kind[j][0]) continue; int s=dp[i-kind[j][0]]+kind[j][1]; if(!vis[i-kind[j][0]][j]) s+=kind[j][2]; if(dp[i]<s) { dp[i]=s; flag=j; } } if(flag>=0) { for(int j=0;j<N;j++) { vis[i][j]=vis[i-kind[flag][0]][j]; } vis[i][flag]++; } } int tmp=0; for(int i=1;i<=M;i++) tmp=max(tmp,dp[i]); printf("%d\n",tmp); } return 0; }
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