FZU 2214 Knapsack problem (超大容量背包)
2017-11-03 20:19
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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
<= 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 件物品放入容量为 B 的背包,每件物品都有它的权重和体积,问所能获得的最大权值。思路
一眼看去只是普通的 01背包,然而数据范围让人无法下手。之前我们考虑前 i 件物品装入容量为 v 的背包所能获得的最大价值,换位思考一下,我们考虑前 i 件物品组成价值为 w 的最小体积,于是这道题就可以解出来啦~
AC 代码
#include <iostream> #include<cstring> #include<algorithm> #define IO ios::sync_with_stdio(false);\ cin.tie(0);\ cout.tie(0); using namespace std; typedef long long LL; const int maxn = 1e5+10; const int mod = 998244353; const int inf = 0x3f3f3f3f; LL n,b; LL w[maxn],v[maxn]; LL dp[5100]; void solve() { memset(dp,inf,sizeof(dp)); dp[0] = 0; for(int i=1; i<=n; i++) for(int j=5000; j>=v[i]; j--) dp[j] = min(dp[j],dp[j-v[i]]+w[i]); for(int i=5000; i>=0; i--) if(dp[i]<=b) { cout<<i<<endl; break; } } int main() { IO; int T; cin>>T; while(T--) { cin>>n>>b; for(int i=1; i<=n; i++) cin>>w[i]>>v[i]; solve(); } return 0; }
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