第六届福建省大学生程序设计竞赛-重现赛，Problem C Knapsack problem【大背包】

Problem C Knapsack problem

Accept: 83 Submit: 457

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
<= 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：给你n个物品，你的背包的重量限制是B，输入n个w[i]重量 v[i]价值。求不超过B的情况下，背包内物品的价格最大。

思路：由于B的上限较大，转化下思路就好了。设置dp[i][j]为前i个物品获得价值为j时所需的最小容量。

状态转移 dp[i][j] = min(dp[i-1][j], dp[i][j-val[i]] + w[i])。优化掉一维就可以了。

ac-code：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <vector>
#define INF 0x3f3f3f3f
#define eps 1e-8
#define MAXN (500+10)
#define MAXM (100000)
#define Ri(a) scanf("%d", &a)
#define Rl(a) scanf("%lld", &a)
#define Rf(a) scanf("%lf", &a)
#define Rs(a) scanf("%s", a)
#define Pi(a) printf("%d\n", (a))
#define Pf(a) printf("%.2lf\n", (a))
#define Pl(a) printf("%lld\n", (a))
#define Ps(a) printf("%s\n", (a))
#define W(a) while(a--)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define MOD 1000000007
#define LL long long
#define lson o<<1, l, mid
#define rson o<<1|1, mid+1, r
#define ll o<<1
#define rr o<<1|1
#define PI acos(-1.0)
using namespace std;
int dp[10000];
int w[MAXN], v[MAXN];
int main()
{
int t;Ri(t);
W(t)
{
int n, W, sumv;
scanf("%d%d",&n, &W);
sumv=0; CLR(dp, INF); dp[0]=0;
for(int i=0;i<n;i++)
{
scanf("%d%d",&w[i],&v[i]);
sumv+=v[i];
}
for(int i=0;i<n;i++)
for(int j=sumv;j>=v[i];j--)
dp[j]=min(dp[j],dp[j-v[i]]+w[i]);
for(int j=sumv;j>=0;j--)
{
if(dp[j]<=W)
{
printf("%d\n",j);
break;
}
}
}
return 0;
}