标签：res time only mem The line cpp names eof
Problem DescriptionGiven 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).
InputThe 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[n] <= 5000
All the inputs are integers.
OutputFor each test case, output the maximum value.
Sample Input Sample Output Source
#include<iostream>
#include<cstring>
#include<algorithm>
#define INF 0x3f3f3f3f
using namespace std;
int w[505];
int v[505];
long long dp[5005];
int n,m;
int main(){
int t;
std::ios::sync_with_stdio(false);
cin>>t;
while(t--){
cin>>n>>m;
int sum=0;
for(int i=0;i<n;i++){
cin>>w[i]>>v[i];
sum+=w[i];
}
memset(dp,INF,sizeof(dp));
dp[0]=0;
for(int i=0;i<n;i++){
for(int j=5000;j>=0;j--){///价值
if(j>=v[i]){
dp[j]=min(dp[j],dp[j-v[i]]+w[i]);
dp[j]=dp[j]==0?(dp[j-v[i]]+w[i]):min(dp[j],dp[j-v[i]]+w[i]);
///存重量
}
}
}
int ans=0;
for(int i=0;i<=5000;i++){
// dp[i]=0x3f3f3f3f-dp[i];
// cout<<i<<" "<<dp[i]<<endl;
if(dp[i]<=m&&i>ans){
ans=i;
}
}
cout<<ans<<endl;
}
}
标签：res time only mem The line cpp names eof
原文地址：https://www.cnblogs.com/luowentao/p/8997007.html
