每日三题-Day6-A（FZU 2214 Knapsack problem 01背包）

Problem 2214 Knapsack problem

Accept: 583 Submit: 2225

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

15

题目大意：

就是01背包……

思路：

因为B太大了，如果按一般的01背包做，肯定超时。

我们可以换个思路，把用一定的容量，换取最大的价值，转换成得到一定的价值，花费的最小的容量。

也就是说，我们的dp数组，只需要5000，然后时间复杂度是O(n*sumv)【500*5000】

dp[i]代表价值为i的，最小的容量。

最后找到一个dp[i]<=B,使得i最大。这个i就是答案。

AC代码：

#include<cstdio>#include<algorithm>#include<cstring>using namespace std;long long dp[5005];int w[505];int v[505];int main(){    int n,t;    int q;    scanf("%d",&t);    while(t--)    {        int sum=0;        scanf("%d%d",&n,&q);        for(int i=0;i<n;i++)        {            scanf("%d%d",&w[i],&v[i]);            sum+=v[i];        }        memset(dp,0x3f3f3f,sizeof(dp));        dp[0]=0;        for(int i=0;i<n;i++)        {            for(int j=sum;j>=v[i];j--)            {                dp[j]=min(dp[j],dp[j-v[i]]+w[i]);            }        }        int ans=0;        for(int i=sum;i>=0;i--)        {            ans=i;            if(dp[i]<=q)break;        }        printf("%d\n",ans);    }    return 0;}