poj 1260 Pearls(dp)

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题目:http://poj.org/problem?id=1260 题意:给出几类珍珠,以及它们的单价,要求用最少的钱就可以买到相同数量的,相同(或更高)质量的珍珠. 珍珠的替代必须是连续的,不能跳跃替代(这个不难证明,因为假如用第i+2类去替代第i类珍珠,会使最终的支付价格降低,那么用第i+1类去替代第i类珍珠会使最终的支付价格更加低) #include<iostream> #include<cstdio> #include<cstring> #include<…

POJ 1260 Pearls 简单dp

1.POJ 1260 2.链接:http://poj.org/problem?id=1260 3.总结:不太懂dp,看了题解 http://www.cnblogs.com/lyy289065406/archive/2011/07/31/2122652.html 题意:珍珠,给出需求,单价,要求用最少的钱就可以买到相同数量的,相同(或更高)质量的珍珠. 把握题意,1.输入时,后输入的珍珠价格一定比前面输入的要贵.2.用高质量珍珠替代低质量. #include<iostream> #include…

(线性结构dp )POJ 1260 Pearls

Pearls Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 10558 Accepted: 5489 Description In Pearlania everybody is fond of pearls. One company, called The Royal Pearl, produces a lot of jewelry with pearls in it. The Royal Pearl has its…

POJ 1260 Pearls （斜率DP）题解

思路: 直接DP也能做,这里用斜率DP. dp[i] = min{ dp[j] + ( sum[i] - sum[j] + 10 )*pr[i]} ; k<j<i => dp[j] - dp[k] <pr[i]*( sum[j] - sum[k] ) 再套模板 #include<queue> #include<cstring> #include<set> #include<map> #include<stack> #i…

poj 1260 Pearls 斜率优化dp

这个题目数据量很小,但是满足斜率优化的条件,可以用斜率优化dp来做. 要注意的地方,0也是一个决策点. #include <iostream> #include <cstdio> #include <cstring> using namespace std; const int maxn=1e2+9; int dp[maxn]; int a[maxn],p[maxn],sum[maxn]; int que[maxn]; bool chk1(int i,int j,int…

POJ 1260 Pearls

Pearls Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 6670 Accepted: 3248 Description In Pearlania everybody is fond of pearls. One company, called The Royal Pearl, produces a lot of jewelry with pearls in it. The Royal Pearl has its name…

POJ 1260 Pearls (动规)

Pearls Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 7210 Accepted: 3543 Description In Pearlania everybody is fond of pearls. One company, called The Royal Pearl, produces a lot of jewelry with pearls in it. The Royal Pearl has its name…