FZOJ--2214--Knapsack problem(背包)
2015-12-27 20:09
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Problem 2214 Knapsack problem
Accept: 5 Submit: 8
Time Limit: 3000 mSec Memory Limit : 32768 KB
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).
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.
For each test case, output the maximum value.
1
5 15
12 42 21 14 101 2
15
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)
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
15 15
12 42 21 14 101 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) #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; #define INF 0x3f3f3f struct node { int u,v; }num[10010]; int dp[10010]; int main() { int t; scanf("%d",&t); while(t--) { int m,n; scanf("%d%d",&n,&m); memset(dp,INF,sizeof(dp)); int V=0; for(int i=0;i<n;i++) { scanf("%d%d",&num[i].u,&num[i].v); V+=num[i].v; } dp[0]=0; for(int i=0;i<n;i++) { for(int j=V;j>=num[i].v;j--) { dp[j]=min(dp[j],dp[j-num[i].v]+num[i].u); } } for(int j=V;j>=0;j--) { if(dp[j]<=m) { printf("%d\n",j); break; } } } return 0; }
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