題干：

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

解題報告：

? ? ? ? 因為這題質量太大了，所以把質量當成價值，價值當成重量，dp[ j ]表示，買到價值j的物品所需要的最小質量。跑0-1背包就可以了。貼一道類似題目【nyoj - 860】 又見0-1背包

AC代碼：

#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 ;}

?

總結

以上是生活随笔為你收集整理的【 FZU - 2214 】Knapsack problem（逆向0-1背包）的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。