7-25 0-1 backpack (50 points)

7-25 0-1 knapsack (50 branch )
Given n(n<=100) Kind of items and a backpack . goods i The weight of is wi, Value is vi, The capacity of the backpack is C(C<=1000). ask : How to choose the items in your backpack , To maximize the total value of items in a backpack ? When choosing items to pack , For each item i There are only two options ： To load or not to load . Don't put things i Load multiple times , You can't just load some items i.

Input format :
share n+1 Line input ： First act n Values and c value , Express n Items and Backpack Capacity c; Next n That's ok , There are two data per line , Separate indication control i(1≤i≤n) The weight and value of the items .

Output format :
Output the maximum total value of items in the backpack .

sample input :
Here's a set of inputs . for example ：

5 10
2 6
2 3
6 5
5 4
4 6
sample output :
Here is the corresponding output . for example ：
15

#include<iostream>
using namespace std;
int n;//n Items 
int w[200];// Weight of stored items 
int v[200];//  Value of goods  
int c;// Backpack Capacity 
int m[200][200]; 
int main(){
    
	cin>>n>>c;
	int i,j;
	for(i=0;i<n;i++){
    
		cin>>w[i]>>v[i];
	}
	for(i=n-1;i>=0;i--){
    
		for(j=0;j<=c;j++){
    
			if(j>=w[i]){
    // It can be put in  
				m[i][j]=max(m[i+1][j],m[i+1][j-w[i]]+v[i]); 
			}
			else{
    
				m[i][j]=m[i+1][j];
			}
		}
	}
	cout<<m[0][c];
}