FZU - 2214 Knapsack problem（0 1 背包++）-CFANZ编程社区

FZU - 2214 Knapsack problem

Accept: 837 Submit: 3249
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[n] <= 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

这道题是0-1背包类的问题，但是不同的是这道题的w[i]非常大，最大为1e9,

emmm,先粘一下普通代码：

for(int i=0;i<n;i++)
      for(int j=0;j<=W;j++)
        if(j<w[i])  dp[i+1][j]=dp[i][j];
        else    dp[i+1][j]=max(dp[i][j],dp[i][j-w[i]]+v[i]);
    printf("%d\n",dp[n][W]);

或者：

for(int i=n-1;i>=0;i--)
      for(int j=0;j<=W;j++)
        if(j<w[i])  dp[i][j]=dp[i+1][j];
        else    dp[i][j]=max(dp[i+1][j],dp[i+1][j-w[i]]+v[i]);
      
    printf("%d\n",dp[0][W]);

这上边的代码时间复杂度是O(nW)

如果做这一道题的话，肯定会超时的啊

所以这一道题DP的对象是价值 v[i]

// 这一道题  DP的对象是价值  v[i] 

//#include<bits/stdc++.h>   //CE
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN=505;
const int MAXV=5000+5;
const int INF=0x3f3f3f3f;
int dp[MAXN][MAXV];
int n,b;
int w[MAXN],v[MAXN];
int main()
{
  int t;
  scanf("%d",&t);
  while(t--)
  {
    scanf("%d %d",&n,&b);
    for(int i=0;i<n;i++)  scanf("%d %d",&w[i],&v[i]);
    memset(dp,0x3f,sizeof(dp));
    dp[0][0]=0;
    for(int i=0;i<n;i++)
    {
      for(int j=0;j<=MAXV;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]]+w[i]);
      }
    }
    int res=0;
    for(int i=0;i<=5000;i++)  if(dp[n][i]<=b)  res=i;//这里的 i<=5000 一定要这样 
    printf("%d\n",res);    
  }
  return 0;
}