poj - 1170 - Shopping Offers（状态压缩dp）

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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;}

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