POJ 1113&&HDU 1348】的更多相关文章

传送门: POJ:点击打开链接 HDU:点击打开链接 以下是POJ上的题: Wall Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 29121   Accepted: 9746 Description Once upon a time there was a greedy King who ordered his chief Architect to build a wall around the King's cast…
题意:凸包周长+一个完整的圆周长.因为走一圈,经过拐点时,所形成的扇形的内角和是360度,故一个完整的圆. 模板题,之前写的Graham模板不对,WR了很多发....POJ上的AC代码 #include<iostream> #include<cstdio> #include<algorithm> #include<cstring> #include<set> #include<stdio.h> #include<stdlib.h…
Wall Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 28462   Accepted: 9498 Description Once upon a time there was a greedy King who ordered his chief Architect to build a wall around the King's castle. The King was so greedy, that he wo…
Wall Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 28157   Accepted: 9401 Description Once upon a time there was a greedy King who ordered his chief Architect to build a wall around the King's castle. The King was so greedy, that he wo…
K-th Number Time Limit: 20000MS   Memory Limit: 65536K Total Submissions: 50247   Accepted: 17101 Case Time Limit: 2000MS Description You are working for Macrohard company in data structures department. After failing your previous task about key inse…
Wall Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 33888   Accepted: 11544 Description Once upon a time there was a greedy King who ordered his chief Architect to build a wall around the King's castle. The King was so greedy, that he w…
poj 1251  && hdu 1301 Sample Input 9 //n 结点数A 2 B 12 I 25B 3 C 10 H 40 I 8C 2 D 18 G 55D 1 E 44E 2 F 60 G 38F 0G 1 H 35H 1 I 353A 2 B 10 C 40B 1 C 200Sample Output 21630 prim算法 # include <iostream> # include <cstdio> # include <cstr…
LINK 题意:给出一个简单几何,问与其边距离长为L的几何图形的周长. 思路:求一个几何图形的最小外接几何,就是求凸包,距离为L相当于再多增加上一个圆的周长(因为只有四个角).看了黑书使用graham算法极角序求凸包会有点小问题,最好用水平序比较好.或者用Melkman算法 /** @Date : 2017-07-13 14:17:05 * @FileName: POJ 1113 极角序求凸包 基础凸包.cpp * @Platform: Windows * @Author : Lweleth (…
题目链接:poj 1113   单调链凸包小结 题解:本题用到的依然是凸包来求,最短的周长,只是多加了一个圆的长度而已,套用模板,就能搞定: AC代码: #include<iostream> #include<algorithm> #include<cstdio> #include<cmath> using namespace std; int m,n; struct p { double x,y; friend int operator <(p a,…
Eight POJ - 1077 HDU - 1043 八数码问题.用hash(康托展开)判重 bfs(TLE) #include<cstdio> #include<iostream> #include<queue> #include<cstring> using namespace std; ,,,,,,,,,}; ]; ][]; queue<int> q; //data[][9]存储上一个状态,data[][10]存储到这个状态的操作,dat…