hdu 5410 CRB and His Birthday（动态规划）

题目：

CRB and His Birthday

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 748    Accepted Submission(s): 395

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 integerT, 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

1100 210 2 120 1 1

 

Sample Output

21
Hint
CRB'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

 

Recommend

wange2014

题意：给了N种礼物，总共M块钱，每件礼物W【i】元，买x个第i种礼物可以得到A[I]*x+B[i]个糖果，问最多能得到多少糖果。

思路：看起来像完全背包，但有一些不同，完全背包是第i种物品的价值为P[i]，也就是说买x个第i种物品能得到x*p[i]的价值，而这里是A[I]*x+B[i]，所以要分是否是第一次买这个礼物，如果是第一次，那么增加的价值是A[i]+B[i]，否则增加A[i].设dp[i][j][0]代表有j元钱，并且现在不买第i物品，dp[i][j][1]代表买这个物品，那么dp[i][j][0]=max(dp[i-1][j][0],dp[i-1][j][1]),dp[i][j][1]=max(dp[i][j-w[i]][0]+A[i]+B[i],dp[i][j-w[i]][1]+A[i]).

代码：

#include <cstdlib>#include <cctype>#include <cstring>#include <cstdio>#include <cmath>#include<climits>#include <algorithm>#include <vector>#include <string>#include <iostream>#include <sstream>#include <map>#include <set>#include <queue>#include <stack>#include <fstream>#include <numeric>#include <iomanip>#include <bitset>#include <list>#include <stdexcept>#include <functional>#include <utility>#include <ctime>using namespace std;#define PB push_back#define MP make_pair#define REP(i,x,n) for(int i=x;i<(n);++i)#define FOR(i,l,h) for(int i=(l);i<=(h);++i)#define FORD(i,h,l) for(int i=(h);i>=(l);--i)#define SZ(X) ((int)(X).size())#define ALL(X) (X).begin(), (X).end()#define RI(X) scanf("%d", &(X))#define RII(X, Y) scanf("%d%d", &(X), &(Y))#define RIII(X, Y, Z) scanf("%d%d%d", &(X), &(Y), &(Z))#define DRI(X) int (X); scanf("%d", &X)#define DRII(X, Y) int X, Y; scanf("%d%d", &X, &Y)#define DRIII(X, Y, Z) int X, Y, Z; scanf("%d%d%d", &X, &Y, &Z)#define OI(X) printf("%d",X);#define RS(X) scanf("%s", (X))#define MS0(X) memset((X), 0, sizeof((X)))#define MS1(X) memset((X), -1, sizeof((X)))#define LEN(X) strlen(X)#define F first#define S second#define Swap(a, b) (a ^= b, b ^= a, a ^= b)#define Dpoint  strcut node{int x,y}#define cmpd int cmp(const int &a,const int &b){return a>b;} /*#ifdef HOME    freopen("in.txt","r",stdin);    #endif*/const int MOD = 1e9+7;typedef vector<int> VI;typedef vector<string> VS;typedef vector<double> VD;typedef long long LL;typedef pair<int,int> PII;//#define HOMEint Scan(){int res = 0, ch, flag = 0;if((ch = getchar()) == '-')//判断正负flag = 1;else if(ch >= '0' && ch <= '9')//得到完整的数res = ch - '0';while((ch = getchar()) >= '0' && ch <= '9' )res = res * 10 + ch - '0';return flag ? -res : res;}/*----------------PLEASE-----DO-----NOT-----HACK-----ME--------------------*/int W[1005],A[1005],B[1005];long long int dp[2005][2];int main(){int T;RI(T);while(T--){    int M,N;    RII(M,N);    for(int i=1;i<=N;i++)        RIII(W[i],A[i],B[i]);    MS0(dp);    for(int i=1;i<=N;i++)        for(int m=0;m<=M;m++)        {           dp[m][0]=max(dp[m][0],dp[m][1]);           if(m>=W[i])           dp[m][1]=max(dp[m-W[i]][0]+A[i]+B[i],dp[m-W[i]][1]+A[i]);           else            dp[m][1]=0;        }        printf("%I64d\n",max(dp[M][0],dp[M][1]));}        return 0;}