FZU 2214 ——Knapsack problem——————【01背包的超大背包】

2214 Knapsack problem

Accept: 6    Submit: 9
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

1

5 15

12 4

2 2

1 1

4 10

1 2

 Sample Output

15

 Source

第六届福建省大学生程序设计竞赛-重现赛（感谢承办方华侨大学）

 

 

题目大意：给你T组数据，每组有n个物品，一个背包容量B，每件有体积和价值。问你这个背包容纳的物品最大价值是多少。每个物品只能放入一次背包。

 

解题思路：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。

 

#include<stdio.h>
#include<algorithm>
#include<string.h>
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 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\n",ans);
    }
    return 0;
}