题意：首先给你个w，表示几组货物，然后给你n以及n个数表示价格，10-价格是利润

每次只能取前i个，求最多的利润对应几个货物，不超过10个解

思路：首先是简单的计算最大的利润，然后就是储存所有的解，然后深搜出所有的可能

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <queue>
#include <vector>
#include <set>

using namespace std;

const int MAXN = 100;

int w, val[MAXN][MAXN];
set<int > ans;
vector<int > p[MAXN];

void cal(int n, int sum) {
	if (n == w) {
		ans.insert(sum);
		return;
	}
	for (int i = 0; i < p[n].size(); i++)		
		cal(n+1, sum+p[n][i]);
}

void solve() {
	ans.clear();
	int valMax[MAXN], valSum, pro;
	valSum = 0;
	memset(valMax, 0, sizeof(valMax));
	for (int i = 0; i < w; i++) {
		p[i].clear();
		pro = 0;
		for (int j = 1; j <= val[i][0]; j++) {
			pro += 10 - val[i][j];
			valMax[i] = max(valMax[i], pro);
		}
		valSum += valMax[i];
		if (valMax[i] == 0)
			p[i].push_back(0);
		pro = 0;
		for (int j = 1; j <= val[i][0]; j++) {
			pro += 10 - val[i][j];
			if (valMax[i] == pro)
				p[i].push_back(j);
		}
	}
	cal(0, 0);
	printf("Maximum profit is %d.\n", valSum);
	printf("Number of pruls to buy:");
	int cnt = 0;
	for (set<int>::iterator i = ans.begin(); i != ans.end() && cnt != 10; i++, cnt++) 
		printf(" %d", *i);
	printf("\n");
}

int main() {
	int t = 0;

	while (scanf("%d", &w) != EOF && w) {
		int b;
		for (int i = 0; i < w; i++) {
			scanf("%d", &val[i][0]);
			for (int j = 1; j <= val[i][0]; j++)
				scanf("%d", &val[i][j]);
		}

		if (t)
			printf("\n");
		printf("Workyards %d\n", ++t);
		solve();
	}

	return 0;
}