FOJProblem 2214 Knapsack problem（01背包+变性思维）

http://acm.fzu.edu.cn/problem.php?pid=2214

Accept: 4    Submit: 6
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）

分析:虐的没点脾气，首先B很大，用01背包开数组开不开，然后我就想什么离散化，结果也没搞出来，最后看到题解居然是把价值下的重量，顿时感觉自己弱爆了，可以把价值看作背包容量啊，就是把这个价值装满的最小重量就是那个对应的重量下的最大价值，一点点变性思维就能解决的事，弱爆了

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 using namespace std;
 const int INF = 0x3f3f3f3f;
 long long thing[];
 int w[],v[];
 int main()
 {
     int t,n,B;
     scanf("%d", &t);
     while(t--)
     {
         int sum = ;
         scanf("%d%d", &n,&B);
         for(int i = ; i <= n; i++)
         {
             scanf("%d%d", &w[i], &v[i]);
             sum += v[i];
         }
         memset(thing, INF, sizeof(thing));
         thing[] = ;
         for(int i = ; i <= n; i++)
         {
             for(int j = sum; j >= v[i]; j--)
             {
                 if(thing[j - v[i]] != INF)
                     thing[j] = min(thing[j], thing[j - v[i]] + w[i]);
             }
         }
         for(int i = sum; i >= ; i--)
         {
             if(thing[i] <= B)
             {
                 printf("%d\n",i);
                 break;
             }
         }
     }
     return ;
 }