FOJ 2214 Knapsack problem 第六届福建省大学生程序设计竞赛 C 01背包DP变种

Knapsack problem
Accept: 202    Submit: 817
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）

题意：01背包，但是背包重量非常大。

思路：不能用传统的公式，题目限制价值总和小于5000，所以可以用价值来dp。dp[i]代表价值总和不大于i的重量的最小值

dp[j] = min(dp[j],dp[j-val[i]]+w[i]);

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include <string>
#include <string.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <iomanip>
#include <time.h>
#include <set>
#include <map>
#include <stack>
using namespace std;
typedef long long LL;
const int INF=0x7fffffff;
const int MAX_N=10000;
const long long BIG=0x3fffffffffffffff;

int val[5005];
long long w[509];
long long dp[5009];//价值不大于i的重量的最小值


int T,n,b;
int main(){
    cin>>T;
    while(T--){
        cin>>n>>b;
        for(int i=0;i<n;i++){
            scanf("%I64d%d",&w[i],&val[i]);
        }


        fill(dp,dp+5001,BIG);
        dp[0]=0;
        for(int i = 0;i<n;i++){
            for(int j = 5000;j>=val[i];j--){
                dp[j] = min(dp[j],dp[j-val[i]]+w[i]);
            }
        }

        long long ans=0;
        for(int i=5000;i>=0;i--){
           // cout<<dp[i]<<endl;
            if(dp[i]<=b){
                ans=i;
                break;
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}

/*
1
5 15
12 4
2 2
1 1
4 10
1 2

*/