【 FZU - 2214 】Knapsack problem（逆向0-1背包）

题干：

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

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

Sample Output

15

解题报告：

        因为这题质量太大了，所以把质量当成价值，价值当成重量，dp[ j ]表示，买到价值j的物品所需要的最小质量。跑0-1背包就可以了。贴一道类似题目【nyoj - 860】 又见0-1背包

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
const int INF = 0x3f3f3f3f;
int w[505],v[505];
int dp[10000 + 5];
int n,m; 
int main()
{
  int t;
  scanf("%d",&t); 
  while(t--) {
    scanf("%d%d",&n,&m); 
    int sum = 0;
    for(int i = 1; i<=n; i++) {
      scanf("%d%d",&w[i],&v[i]);
      sum += v[i];
    }
    memset(dp,INF,sizeof(dp));
    dp[0] = 0;
    for(int i = 1; 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--) {
      if(dp[i] <= m) {
        ans = i;break;
      } 
    }
    printf("%d\n",ans);
  }
  return 0 ;
 }