HDOJ 1712 ACboy needs your help(分组背包入门)

题意:

给出课程 i 所付出时间 j 所取得的收益,求最大收益。

思路:

分组背包入门:dp[i][j] = max(dp[i-1][j-k] + valk)  dp[i][j] 代表前 i 种物品付出时间 j 所得到的最大收益。

#include <iostream>
#include <algorithm>
using namespace std;

const int MAXN = 110;
const int MAXD = 110;

int dp[MAXD], A[MAXN];

int main()
{
    int n, m;
    while (scanf("%d %d", &n, &m) && n && m)
    {
        memset(dp, 0, sizeof(dp));
        for (int i = 0; i < n; ++i)
        {
            for (int j = 1; j <= m; ++j)
                scanf("%d", &A[j]);

            for (int v = m; v > 0; --v)
                for (int j = 0; j <= v; ++j)
                    dp[v] = max(dp[v], dp[v - j] + A[j]);
        }
        int ret = 0;
        for (int v = 1; v <= m; ++v)
            ret = max(ret, dp[v]);

        printf("%d\n", ret);
    }
    return 0;
}
原文地址:https://www.cnblogs.com/kedebug/p/2913841.html