超大背包问题

有一类背包问题中背包的体积很大但是价值很小，这种背包中如果按照正常的思维来写的话很容易超内存，可以换种思维，dp[I]表示价值为I的价值用的最小空间；

例题：

Knapsack problem

Crawling in process... Crawling failed  

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

  1
  5 15
12 4
  2 2
1 1
4 10
1 2

Sample Output

  15

解体思路：首先这个很清楚看出来是个01背包。但是这个背包容量特别大，同时物品的体积很大，总的价值却很少。寻常意义上dp[i]表示背包容量为i时的最大价值，
那么我们把这个dp[]数组的含义改变一下，dp[i]表示装价值为i时所需的最小容量。

感悟：唉，压线过得真持基；
#include
#include
#include
using namespace std;
const int maxn = 550;
const int INF = 0x3f3f3f3f;
int w[maxn], v[maxn];
int dp[5500];
int main(){
    freopen("in.txt","r",stdin);
    int T, n, B;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&n,&B);
        int V = 0;
        for(int i = 1; i <= n; i++){
            scanf("%d%d",&w[i],&v[i]);
            V += v[i];
        }
        memset(dp,INF,sizeof(dp));
        dp[0] = 0;
        for(int i = 1; i <= n; i++){
            for(int j = V; j >= v[i]; j--){
                dp[j] = min(dp[j],dp[j-v[i]]+w[i]);
             //   printf("%d ",dp[j]);
            }
        }
        int ans = 0;
        for(int i = V; i >= 0; i--){
            if(dp[i] <= B){
                ans = i; break;
            }
        }
        printf("%d ",ans);
    }
    return 0;
}