fzu 2214 Knapsack problem【超大01背包】

fzu 2214 Knapsack problem【超大01背包】

题目来戳呀
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
1
5 15
12 4
2 2
1 1
4 10
1 2
Sample Output
15
题意：n个背包，有w和v，求在容量为b的背包中的最大价值。但是物品质量特别特别大！
想法：01背包问题，但是质量大容量大所以不能直接上模板。在寻常意义上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。

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn = 550;
const int INF = 0x3f3f3f3f;
int w[maxn], v[maxn];
int dp[5500];
int main()
{int T, n, B;scanf("%d",&T);while(T--){scanf("%d%d",&n,&B);int sumv = 0;for(int i = 1; i <= n; i++){scanf("%d%d",&w[i],&v[i]);sumv += v[i];}memset(dp,INF,sizeof(dp));dp[0] = 0;for(int i = 1; i <= n; i++){//对于i个物体，遍历所有价值jfor(int j = sumv; j >= v[i]; j--){dp[j] = min(dp[j],dp[j-v[i]]+w[i]);，//前者放，后者不放}}int ans = 0;for(int i = sumv; i >= 0; i--){if(dp[i] <= B){//遍历价值由大到小的质量与容量比较ans = i;break;}}printf("%dn",ans);}return 0;
}