Nyoj 311 full backpack

Full backpack time limit: 3000 MS | memory limit: 65535 kb difficulty: 4

Description

A full backpack defines n items and a backpack with a capacity of V. Each item has an unlimited number of items available. The volume of item I is C and the value is W. Solving which items are loaded into a backpack can make the total size of these items not exceed the total size of the backpack, and the total value is the largest. This question requires that when a backpack is filled with a backpack, find the sum of the maximum value. If it cannot be filled with a backpack, output No

 

Input

Row 1: N indicates the number of groups of test data (n <7 ).
Next, the first row of test data in each group contains two integers, M and V. M indicates the number of item types, and V indicates the total size of the backpack. (0 <m <= 2000,0 <v <= 50000)
Each row in the next m row has two integers C, W representing the weight and value of each item respectively (0 <C <100000, 0 <W <)

Output

Corresponding to the output results of each group of test data (if it can be filled with a backpack, the maximum value of the items in the backpack is output when it is filled with a backpack. If it cannot be filled with a backpack, output No)

Sample Input

21 52 22 52 25 1

Sample output

NO1

Uploaded

ACM _ Zhao minghao

Problem solving: RT

 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 #include <vector> 6 #include <climits> 7 #include <algorithm> 8 #include <cmath> 9 #define LL long long10 using namespace std;11 const int INF = INT_MAX>>2;12 int c[2001],w[2001],dp[50001];13 int main(){14     int kase,n,i,j,v,k;15     scanf("%d",&kase);16     while(kase--){17         scanf("%d %d",&n,&v);18         for(i = 1; i <= n; i++)19             scanf("%d %d",c+i,w+i);20         for(i = 0; i <= v; i++)21             dp[i] = -INF;22         dp[0] = 0;23         for(i = 1; i <= n; i++){24             for(j = c[i]; j <= v; j++)25                 if(dp[j] < dp[j-c[i]]+w[i]) dp[j] = dp[j-c[i]]+w[i];26         }27         if(dp[v] > 0){28             printf("%d\n",dp[v]);29         }else puts("NO");30     }31     return 0;32 }

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