题链: http://poj.org/problem?id=1259 题解: 计算几何,凸包,DP 题意:给出N($N\leq100$)个点,求出最大的凸包使得凸包里面不存在点(边上可以有).输出最大凸包的面积. 把所有点按x从小到大排序(若x相同则按y从小到大排序). 依次枚举每个点p,把它作为凸包的左下角来计算此时可以形成的最大凸包. 做法如下: 将p号点作为原点,取出p+1~N的点(共m=N-p个点),对这m个点按极角逆时针排序. 然后定义DP状态: f[j][i]:表示以i,j作为当前凸…

题目:http://poj.org/problem?id=1258 题意:模板题 和2485差不多 就是求相连后的最小值. #include <iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<stack> #include<queue> #include<cmath> #include<algorithm> usin…

[题意] 有n个巨人要去Park聚会.巨人A和先到巨人B那里去,然后和巨人B一起去Park.B君是个土豪,他家的停车场很大,可以停很多车,但是Park的停车场是比较小.只能停k辆车.现在问你在这个限制条件下.巨人到达Park的最短距离. 如果把那个条件去掉.那么就是就是求最小生成树.加上那个条件其实就是求顶点度数限制为k的最小生成树. Input Input will consist of one problem instance. The first line will contain a s…

Cow Picnic Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 4928   Accepted: 2019 Description The cows are having a picnic! Each of Farmer John's K (1 ≤ K ≤ 100) cows is grazing in one of N (1 ≤ N ≤ 1,000) pastures, conveniently numbered…

The Contortion Brothers are a famous set of circus clowns, known worldwide for their incredible ability to cram an unlimited number of themselves into even the smallest vehicle. During the off-season, the brothers like to get together for an Annual C…

Picnic Planning Time Limit: 5000MS   Memory Limit: 10000K Total Submissions:11615   Accepted: 4172 Description The Contortion Brothers are a famous set of circus clowns, known worldwide for their incredible ability to cram an unlimited number of them…

Description The Contortion Brothers are a famous set of circus clowns, known worldwide for their incredible ability to cram an unlimited number of themselves into even the smallest vehicle. During the off-season, the brothers like to get together for…

题目链接:http://poj.org/problem?id=1639 题意: 给你一个无向图,n个节点,m条边,每条边有边权. 让你求一棵最小生成树,同时保证1号节点的度数<=k. 题解: 最小度限制生成树: (1)不用与1号节点相连的边,跑一次kruskal,得到了deg个连通块. (2)选取与1相连的deg条边,并使得边尽可能小,将1与这些连通块连起来,得到了一棵deg度最小生成树. (3)利用当前的deg度最小生成树,求出deg+1度最小生成树.如此重复至k度最小生成树: I. 在当前生…

[题目链接] http://poj.org/problem?id=1639 [算法] 首先,我们可以用深度优先遍历求出1号节点去除后有几个联通块 设共有T个联通块,若T > K则无解,否则 : 求出这些联通块的最小生成树,得到一棵最小T度生成树,我们尝试改动(K - T)条边,使得答案变得更小,具体过程如下 : 枚举所有与1相连的边,若这条边不在当前的生成树中,我们用广度优先搜索求出生成树上1号节点到该条边的节点中最长的边,用这条边的权值减去枚举边的权值即为生成树权值和变小了多少,求出这个变小的…

https://vjudge.net/problem/POJ-1639 题意: 有一群人,他们要去某一个地方,每个车可以装无数个人,给出了n条路,包含的信息有路连接的地方,以及路的长度,路是双向的,但是终点只有一个,并且终点能停的车的数量是有限制的,问最少走的路是多少. 思路: 因为终点的停车的数量是有限制的,所以终点的度是有限制的,又因为这题可以用最小生成树解决,所以就是度限制最小生成树. 度限制最小生成树求解思想并不复杂,首先我们把有度限制的点给忽略,然后给每一个连通分量求最小生成树,最后把…