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Problem DescriptionGiven 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).
InputThe 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.
OutputFor each test case, output the maximum value.
Sample Input Sample Output Source#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
const int maxn = 550;
const int INF = 0x3f3f3f3f;
int w[maxn], v[maxn];
int dp[5500];
int main(){
int T, n, B;
scanf("%d",&T);
while(T--){
scanf("%d%d",&n,&B);
int V = 0;
for(int i = 1; i <= n; i++){
scanf("%d%d",&w[i],&v[i]);
V += v[i];
}
memset(dp,INF,sizeof(dp));
dp[0] = 0;
for(int i = 1; i <= n; i++){
for(int j = V; j >= v[i]; j--){
dp[j] = min(dp[j],dp[j-v[i]]+w[i]);
// printf("%d ",dp[j]);
}
}
int ans = 0;
for(int i = V; i >= 0; i--){
if(dp[i] <= B){
ans = i; break;
}
}
printf("%d\n",ans);
}
return 0;
}
FZU 2214 ——Knapsack problem——————【01背包的超大背包】
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原文地址：http://www.cnblogs.com/chengsheng/p/5080860.html
