题意：b(0 <= b <= 5)种物品，每种有个标号c(1 <= c <= 999)，有个需要购买的个数k(1 <= k <=5)，有个单价p(1 <= p <= 999)，有s(0 <= s <= 99)种组合优惠方案，问完成采购最少需要多少钱。
题目链接：http://poj.org/problem?id=1170
——>>已有b种物品，再将每种优惠分别看成一种新物品，剩下就是完全背包问题了。。
设dp[i]表示购买状态为 i 时的最少花费（关于购买状态：00032表示第0种物品买2个，第1种物品买3个），则状态转移方程为：
dp[i + product[j].nState] = min(dp[i + product[j].nState], dp[i] + product[j].nPrice)（j是枚举各种物品的物品下标）;
#include <cstdio>
#include <cstring>
#include <algorithm>using std::min;const int MAXN = 5 + 1;
const int MAXS = 6 * 6 * 6 * 6 * 6;
const int MAX_SIX = 6;
const int MAX_ID = 999 + 1;
const int MAX_OFFER = 99 + 1;struct PRODUCT
{int nId;int nNum;int nPrice;int nState;
} product[MAXN + MAX_OFFER];int nType;
int dp[MAXS];
int nTargetState;
int nSixPow[MAX_SIX];
int nId2Bit[MAX_ID];void Init()
{nType = 0;nTargetState = 0;
}void GetSixPow()
{nSixPow[0] = 1;for (int i = 1; i < MAX_SIX; ++i){nSixPow[i] = nSixPow[i - 1] * 6;}
}void CompletePack()
{memset(dp, 0x3f, sizeof(dp));dp[0] = 0;for (int i = 0; i < nTargetState; ++i){for (int j = 0; j < nType; ++j){if (i + product[j].nState > nTargetState) continue;dp[i + product[j].nState] = min(dp[i + product[j].nState], dp[i] + product[j].nPrice);}}printf("%d\n", dp[nTargetState]);
}int main()
{int b, s, n;GetSixPow();while (scanf("%d", &b) == 1){Init();for (int i = 0; i < b; ++i){scanf("%d%d%d", &product[i].nId, &product[i].nNum, &product[i].nPrice);product[i].nState = nSixPow[i];nTargetState += nSixPow[i] * product[i].nNum;nId2Bit[product[i].nId] = i;}scanf("%d", &s);for (int i = 0; i < s; ++i){int nId, nCnt;product[b + i].nState = 0;scanf("%d", &n);for (int j = 0; j < n; ++j){scanf("%d%d", &nId, &nCnt);product[b + i].nState += nSixPow[nId2Bit[nId]] * nCnt;}scanf("%d", &product[b + i].nPrice);}nType = b + s;CompletePack();}return 0;
}