uva10130解题报告

Root :: Contest Volumes :: Volume CI
10130 - SuperSale Time limit: 3.000 seconds

Root :: Contest Volumes :: Volume CI

SuperSale

There is a SuperSale in a SuperHiperMarket. Every person
can take only one object of each kind, i.e. one TV, one carrot, but
for extra low price. We are going with a whole family to that
SuperHiperMarket. Every person can take as many objects, as he/she
can carry out from the SuperSale. We have given list of objects
with prices and their weight. We also know, what is the maximum
weight that every person can stand. What is the maximal value of
objects we can buy at SuperSale?

Input Specification

The input consists of T test
cases. The number of them
(1<=T<=1000) is given on
the first line of the input file.

Each test case begins with a line
containing a single integer number N that indicates the
number of objects (1 <= N <=
1000). Then followsN lines, each containing two
integers: P and W. The first integer
(1<=P<=100) corresponds to the price
of object. The second integer
(1<=W<=30) corresponds to the weight
of object. Next line contains one integer
(1<=G<=100) it’s the number of people
in our group. Next G lines contains maximal weight
(1<=MW<=30) that can stand this
i-th person from our family
(1<=i<=G).

Output Specification

For every test case your program has to
determine one integer. Print out the maximal value of goods which
we can buy with that family.

Sample Input

2

3

72 17

44 23

31 24

1

26

6

64 26

85 22

52 4

99 18

39 13

54 9

4

23

20

20

26

Output for the Sample Input

72
514
这题就是简单的0-1背包Every person can take only one
object of each kind这句很关键（提醒就是0-1背包）。。
老是被0-1背包的模版束缚，不知道变通。不知道这个条件的作用Next line
contains one integer (1<=G<=100) it’s
the number of people in our group. Next G lines contains maximal
weight (1<=MW<=30) that can stand
this i-th person from our family
(1<=i<=G).不知道怎么处理它。
Print out the maximal value of goods which
we can buy with that
family.在看这句就差不多明白了。。有多个人我们要做的就是在背包九讲中0-1背包的基础上累加每一个人的max值就ok了。头脑不会转，没办法啊。。多思少巧。
#include<stdio.h>

#include<string.h>

#define maxn 1005

int f[10001],p[maxn],w[maxn];

int c,ans,T,N,G;

int max(int a,int b)

{

return a>b?a:b;

}

int main()

{

scanf("%d",&T);

while(T--)

{

scanf("%d",&N);

int i,j;

for(i=1;i<=N;i++)

scanf("%d
%d",&p[i],&w[i]);

scanf("%d",&G);

ans=0;

while(G--)

{

scanf("%d",&c);

memset(f,0,sizeof(f));

for(i=1;i<=N;i++)

{

for(j=c;j>=0;j--)

{

if(j>=w[i])

f[j]=max(f[j],f[j-w[i]]+p[i]);

}

}

ans+=f[c];//累加；

}

printf("%d\n",ans);

}

return 0;

}