FZU - 2214 Knapsack problem (dp（对价值dp）)

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

1
5 15
12 4
2 2
1 1
4 10
1 2

Sample Output

15

Hint

通过题中的数据范围可知，，对w dp是不可行的。。。

对vdp

这就很尴尬了。。。

(1-n)dp[i] = inf;

dp[0]=0;

dp[v]=w;

开始的时候除了0价值的所需要重量是0

其他的价值都是无穷大

让后我们开始dp

从第一个开始。

dp[j]=min(dp[j],dp[j-v[i]]+w[i]);

每一个价值选取重量最小的价值更新上去

（从后往前dp这是01背包）（从前往后是完全背包）

（这里需要的是01 背包）

for（int i=0;i<n;i++)

{

for(int j=max_v;j>=v[i];j--)

{

dp[j]=min(dp[j],dp[j-v[i]]+w[i]);

}

}

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define N 500*1000000000
int w[10003];
int v[10003];
long long dp[5005];
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int n,m;
scanf("%d%d",&n,&m);
int sv=0;
for(int i=0; i<n; i++)
{
scanf("%d%d",&w[i],&v[i]);
sv+=v[i];
}
dp[0]=0;
for(int i=1;i<=sv;i++)
{
dp[i]=N;
}
for(int i=0;i<n;i++)
{
for(int j=sv;j>=v[i];j--)
{
dp[j]=min(dp[j],dp[j-v[i]]+w[i]);
}
}
for(int i=sv;i>=0;i--)
{
if(dp[i]<=m)
{
printf("%d\n",i);
break;
}
}
}
}