POJ - 1170 Shopping Offers (五维DP)
题目大意:有一个人要买b件商品,给出每件商品的编号,价格和数量,恰逢商店打折。有s种打折方式。问怎么才干使买的价格达到最低
解题思路:最多仅仅有五种商品。且每件商品最多仅仅有5个,所以能够用5维dp来表示。每一个维度都代表一件商品的数量
打折的方式事实上有b + s种。将每种商品单件卖的也算一种打折方式
这题有个坑点,就是b或者s有可能为0
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
using namespace std;
#define maxn 6
#define maxs 200
#define INF 0x3f3f3f3f
struct product{
int c, k, p;
}pro[maxn];
struct offer{
int num[maxn], price;
}off[maxs];
int b, s;
int dp[maxn][maxn][maxn][maxn][maxn];
map<int,int> m;
void init() {
memset(dp, -1, sizeof(dp));
memset(off, 0, sizeof(off));
m.clear();
for(int i = 0; i < b; i++) {
scanf("%d%d%d", &pro[i].c, &pro[i].k, &pro[i].p);
m[pro[i].c] = i;
off[i].num[i] = 1;
off[i].price = pro[i].p;
}
if(b < 5) {
for(int i = b; i < 5; i++)
pro[i].k = pro[i].p = 0;
}
scanf("%d", &s);
int n, c, k;
for(int i = b; i < b + s; i++) {
scanf("%d", &n);
for(int j = 0; j < n; j++) {
scanf("%d%d", &c, &k);
off[i].num[m[c]] += k;
}
scanf("%d", &off[i].price);
}
}
int dfs(int a1, int a2, int a3, int a4, int a5) {
if(dp[a1][a2][a3][a4][a5] != -1)
return dp[a1][a2][a3][a4][a5];
dp[a1][a2][a3][a4][a5] = INF;
for(int i = 0; i < b + s; i++) {
if(a1 >= off[i].num[0] && a2 >= off[i].num[1] && a3 >= off[i].num[2] && a4 >= off[i].num[3] && a5 >= off[i].num[4])
dp[a1][a2][a3][a4][a5] = min(dp[a1][a2][a3][a4][a5], off[i].price + dfs(a1-off[i].num[0],a2-off[i].num[1],a3-off[i].num[2],a4-off[i].num[3],a5-off[i].num[4]));
}
return dp[a1][a2][a3][a4][a5];
}
void solve() {
dp[0][0][0][0][0] = 0;
if(b == b + s) {
int t = 0;
for(int i = 0; i < 5; i++)
t += pro[i].p * pro[i].k;
printf("%d\n", t);
}
else
printf("%d\n",dfs(pro[0].k, pro[1].k, pro[2].k, pro[3].k, pro[4].k));
}
int main() {
scanf("%d", &b);
init();
solve();
return 0;
}