Fuzhou oj--福建省校赛重现

Problem C Knapsack problem

Accept: 83 Submit: 457

Time Limit: 3000 mSec Memory Limit : 32768 KB

Problem Description

Given a set of n items, each with a weight w[i] and a value v[i], determine a way to choose the items into a knapsack so that the total weight is less than or equal to a given limit B and the total value is as large as possible. Find the maximum total value.
(Note that each item can be only chosen once).

Input

The first line contains the integer T indicating to the number of test cases.

For each test case, the first line contains the integers n and B.

Following n lines provide the information of each item.

The i-th line contains the weight w[i] and the value v[i] of the i-th item respectively.

1 <= number of test cases <= 100

1 <= n <= 500

1 <= B, w[i] <= 1000000000

1 <= v[1]+v[2]+...+v
<= 5000

All the inputs are integers.

Output

For each test case, output the maximum value.

Sample Input

15 1512 42 21 14 101 2

Sample Output

15

解体思路：0-1背包，只不过不是以背包容量作为限制，而是以价值作为选择限制。在输入时定义变量为 long long竟然wa了，最后改为int才过，无语.......

代码如下：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int w[510];
int v[510];
long long f[5000+10];
int main(){
int t,sum,n,b;
scanf("%d",&t);
while(t--){
sum=0;
scanf("%d%d",&n,&b);
for(int i=1;i<=n;i++){
scanf("%d%d",&w[i],&v[i]);
sum+=v[i];
}
memset(f,0x3f,sizeof(f));
f[0]=0;
for(int i=1;i<=n;i++){
for(int j=sum;j>=v[i];j--){
f[j]=min(f[j],f[j-v[i]]+w[i]);
}
}
for(int i=sum;i>=0;i--){
if(f[i]<=b){
printf("%d\n",i);
break;
}
}
}
return 0;
}