Description
Hot dog is a very delicious food，and Goc like it very much. The picture below shows how much does Goc like it. hhhhh...
Given a set of hot dog, each with a price and a happy value , determine a way to choose the items into a knapsack so that the total price is less than or equal to a given limit and the total happy value is as large as possible. Find the maximum total happy value. (Note that each item can be only chosen once).
Input
The first line contains the integer indicating to the number of test cases.
For each test case, the first line contains the integers and .
Following lines provide the information of each item.
The line contains the price and the happy value of the item respectively.
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
#include <cstdio>
#include <iostream>
#include <queue>
#include <cstring>
#include <
string>
#include <cmath>
using namespace std;
#define ll long long
int t, n, pri, p[
500+
8], v[
500+
8], dp[
500+
8][
5000+
8], sum;
void solve()
{
fill(dp[0], dp[
0] +
5005,
0x3f3f3f3f);
dp[0][
0] =
0;
for(
int i =
0; i<n; i++
)
{
for(
int j =
0; j <=
5001; j++
)
{
if(j<v[i]) dp[i+
1][j] =
dp[i][j];
else dp[i+
1][j] = min(dp[i][j],dp[i][j-v[i]]+
p[i]);
}
}
int miao =
0;
for(
int i =
0; i <=
5001; i++
)
{
if(dp[n][i] <= pri) miao =
i;
}
printf("%d\n", miao);
}
int main()
{
scanf("%d", &
t);
memset(dp, 0,
sizeof(dp));
memset(p, 0,
sizeof(p));
memset(v, 0,
sizeof(v));
while(t--
)
{
scanf("%d%d", &n, &
pri);
for(
int i =
0; i<n; i++
)
{
scanf("%d%d", &p[i], &
v[i]);
}
solve();
}
return 0;
}
转载于:https://www.cnblogs.com/RootVount/p/10326113.html
相关资源：JAVA上百实例源码以及开源项目
