这个题目难度非常大,首先对于老师的一种方案,应用分数规划的一般做法,求出所有的c=t-rate*p,如果没有选择的c值中的最大值比选择了的c值中的最小值大,那么这个解是可以改进的. 那么问题就转化成了怎么求最小的c和最大的c. t-rate*p 求这种类型的最值,并且rate是单调的,那么就可以考虑利用斜率优化的那种办法来维护决策点. 考虑两个决策点,得到ti-tj>rate(pi-pj)  但是这个pi pj的大小不能确定,我们知道可以利用斜率优化的问题不仅仅要rate单调,还需要pi 单调…

Best Cow Fences Time Limit: 1000MS Memory Limit: 30000K Description Farmer John's farm consists of a long row of N (1 <= N <= 100,000)fields. Each field contains a certain number of cows, 1 <= ncows <= 2000. FJ wants to build a fence around a…

http://172.20.6.3/Problem_Show.asp?id=1636 复习了prim,分数规划大概就是把一个求最小值或最大值的分式移项变成一个可二分求解的式子. #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<cstdlib> using namespace std; ;…

思路:以val[u]-ans*edge[i].len最为边权,判断是否有正环存在,若有,那么就是ans小了.否则就是大了. 在spfa判环时,先将所有点进队列. #include<iostream> #include<cstdio> #include<queue> #include<cstring> #include<algorithm> #include<cmath> #define Maxn 1010 #define Maxm 6…

果然比二分要快将近一倍.63MS.二分94MS. #include <iostream> #include <algorithm> #include <cstdio> #include <ctime> #include <cstdlib> #include <cmath> using namespace std; const int maxn=1005; double a[maxn],b[maxn]; const double eps…

题目链接:http://poj.org/problem?id=3621 Sightseeing Cows Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 11526   Accepted: 3930 Description Farmer John has decided to reward his cows for their hard work by taking them on a tour of the big ci…

Desert King Time Limit: 3000MS   Memory Limit: 65536K Total Submissions:29775   Accepted: 8192 Description David the Great has just become the king of a desert country. To win the respect of his people, he decided to build channels all over his count…

Dropping tests Time Limit: 1000MS   Memory Limit: 65536K Total Submissions:17069   Accepted: 5925 Description In a certain course, you take n tests. If you get ai out of bi questions correct on test i, your cumulative average is defined to be . Given…

P2877 [USACO07JAN]牛校Cow School 01分数规划是啥(转) 决策单调性分治,可以解决(不限于)一些你知道要用斜率优化却不会写的问题 怎么证明?可以暴力打表 我们用$ask(l,r,dl,dr)$表示处理区间$[l,r]$时,这段区间的决策点已固定在$[dl,dr]$中 设$mid=(l+r)/2$,暴力处理$mid$的最优决策点$dm$ 再向下分治$ask(l,mid-1,dl,dm)$,$ask(mid+1,r,dm,dr)$ 对于本题,先按$t[i]/p[i]$从大…

http://poj.org/problem?id=2728 题意: 在这么一个图中求一棵生成树,这棵树的单位长度的花费最小是多少? 思路: 最优比率生成树,也就是01分数规划,二分答案即可,题目很简单,因为这题是稠密图,所以用prim算法会好点. #include<iostream> #include<algorithm> #include<cstring> #include<cstdio> #include<vector> #include&…