动态规划01背包问题

01背包问题，用动态规划来解

在对第i个物品检索时随时更新如果有j的空余位置，能装下的最大价值为多少

#include <iostream>//#include <stdio>#include <cstdio>#include <algorithm>#include <vector>#include <cstring>#include <map>#include <cmath>#include <string>#include <queue>#include <stack>using namespace std;const int N = 2000;int max(int a, int b){if (a > b)return a;elsereturn b;}int best[N][N];int value[N];int need[N];int n, v, w;int main(){int i, j;while (scanf("%d", &n) != EOF){scanf("%d", &w);for (i = 1; i <= n; i++){scanf("%d%d", &need[i], &value[i]);//cin >> value[N] >> need[N];}for (i = 0; i <= w; i++)best[0][i] = 0;//下面是核心代码for(i=1;i<=n;i++)//对每个物品进行一遍检索for (j = 1; j <= w; j++)//假设有j的空余位置，可以有多大的最大价值{if (j < need[i])best[i][j] = best[i - 1][j];elsebest[i][j] = max(best[i - 1][j - need[i]] + value[i], best[i - 1][j]);//状态转移方程}//for (i = 0; i <= n ; i++)//for(j=0;j<=w;j++)//cout <<"i= "<<i<<" ,j= "<<j<<"   "<< best[i][j] << endl;cout << best[n][w] << endl;}return 0;}