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
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 using namespace std;
 6 const int INF = 0x3f3f3f3f;
 7 long long thing[5500];
 8 int w[550],v[550];
 9 int main()
10 {
11     int t,n,B;
12     scanf("%d", &t);
13     while(t--)
14     {
15         int sum = 0;
16         scanf("%d%d", &n,&B);
17         for(int i = 1; i <= n; i++)
18         {
19             scanf("%d%d", &w[i], &v[i]);
20             sum += v[i];
21         }
22         memset(thing, INF, sizeof(thing));
23         thing[0] = 0;
24         for(int i = 1; i <= n; i++)
25         {
26             for(int j = sum; j >= v[i]; j--)
27             {
28                 if(thing[j - v[i]] != INF)
29                     thing[j] = min(thing[j], thing[j - v[i]] + w[i]);
30             }
31         }
32         for(int i = sum; i >= 0; i--)
33         {
34             if(thing[i] <= B)
35             {
36                 printf("%d\n",i);
37                 break;
38             }
39         }
40     }
41     return 0;
42 }
View Code

 

posted @ 2015-12-27 20:00  zhaop  阅读(346)  评论(0编辑  收藏  举报