01揹包

設f[i][j]表示考慮完前i個物體，揹包的容量爲j（物體的總體積小於等於j）時的最大價值

答案就是f[n][m]

邊界f[0][...] = 0

轉移：決策一：選f[i][j]->f[i+1][j+weight[i+1]]

         不選f[i][j]->f[i+1][j];

代碼：

#include<iostream>
#include<cmath>
#include<iomanip>
#include<algorithm>
#include<cstdio>

using namespace std;

const int MAXN = 10000;

int n, m;//n爲物品數量  m爲揹包容量 m <= 200 n <= 30
int Wi[MAXN], Ci[MAXN]; //Wi重量，Ci價值；//int Wi, Ci;
int laji[MAXN][MAXN];//int  laji[MAXN];
int main( )
{
	scanf("%d %d", &m, &n);
	for(int i = 1;i <= n; i++)
	{
		scanf("%d %d", &Wi[i], &Ci[i]);
	}
	for(int i = 1;i <= n;i++)
	{
		for(int j = 0;j <= m;j++)
		{
			laji[i][j] = laji[i - 1][j];
			if(j >= Wi[i])
			{
				laji[i][j] = max(laji[i - 1][j], laji [i - 1][j - Wi[i]] + Ci[i]);
			}
			
		}
	}
	
	printf("%d", laji[n][m]);
	
}