题意:给n个点,可以将每个点的x,y的欧几里得距离(就是坐标系里两点距离公式)看作距离,z的差值即为费用差,求的是所有最小生成树中的min(边费用和/边距离和). 思路:其实挑战P143有类似的列题,用的是二分枚举答案的方法,只不过不是树.这一题仅仅需要将题给图找出最小生成树,然后同样枚举即可. 虽然网上有许多高级的名词什么最优比率xxx之类的..以及迭代的方法,不过我认为用二分也很好,易于想到也可以加深理解. #include <iostream> #include <string&g…

Desert King Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 25310   Accepted: 7022 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 coun…

依然是学习分析方法的一道题 求一个长度为n的序列中的一个平均值最大且长度不小于L的子段,输出最大平均值 最值问题可二分,从而转变为判定性问题:是否存在长度大于等于L且平均值大于等于mid的字段和 每个数与mid作差再转变为求非负子段 子段和问题应该利用前缀和C,长度大于等于L的字段和最大值可表示为 max{Aj+1 + Aj+2 ... + Ai},i-j+1-1>=L,j+1>=1 等价于 max{Ci-min{Cj}},L<=i<=n,0<=j<=i-L 注意是i=…

tree 时间限制: 3 Sec  内存限制: 512 MB 题目描述 给你一个无向带权连通图,每条边是黑色或白色.让你求一棵最小权的恰好有need条白色边的生成树. 题目保证有解. 输入 第一行V,E,need分别表示点数,边数和需要的白色边数. 接下来E行,每行s,t,c,col表示这边的端点(点从0开始标号),边权,颜色(0白色1黑色). 输出 一行表示所求生成树的边权和. V<=50000,E<=100000,所有数据边权为[1,100]中的正整数. 样例输入 2 2 1 0 1 1…

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; ;…

题目链接 \(Description\) 将n个村庄连成一棵树,村之间的距离为两村的欧几里得距离,村之间的花费为海拔z的差,求花费和与长度和的最小比值 \(Solution\) 二分,假设mid为可行的某一生成树的解,则应有 \((∑cost)/(∑dis) = mid\) 变形得 \(\sum(cost-mid*dis) = 0\) 取cost-mid*dis为边权,Prim求最小生成树(即尽可能满足mid) 若\(\sum(cost-mid*dis) > 0\),说明怎么也满足不了mid,m…

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 country to bring water to every village. Villages which are connected to his capital village will be watered. As…

别人解题报告的链接: http://blog.sina.com.cn/s/blog_691190870101626q.html 说明一下关于精度的问题,当结果是精确到小数点后3为,你自然要把误差定为至少10^(-4),我定的是10^(-8)````这里多定点没事的··· 然后对于POJ上的提交,如果是用C++提交,可以写printf("%.3lf\n",ans); 但是如果是用的G++提交,就得用printf("%.3f\n",ans); 当然ans定义的是doub…

Desert King http://poj.org/problem?id=2728 Time Limit: 3000MS   Memory Limit: 65536K       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 country t…

题目: http://poj.org/problem?id=2728 题解: 二分比率,然后每条边边权变成w-mid*dis,用prim跑最小生成树就行 #include<cstdio> #include<algorithm> #include<cstring> #include<cmath> #define N 1005 using namespace std; int n,tot; double x[N],y[N],z[N],dis[N]; bool v…