POJ 1862 Stripies】的更多相关文章

原题链接:http://poj.org/problem?id=1862 简单题,贪心+优先队列主要练习一下stl大根堆 写了几种实现方式写成类的形式还是要慢一些... 手打的heap: 1: #include<cstdio> #include<cstdlib> #include<cmath> #include<iostream> class Solution{ public: ; int sz; double heap[Max_N]; inline void…
http://poj.org/problem?id=1862 题目大意: 有一种生物能两两合并,合并之前的重量分别为m1和m2,合并之后变为2*sqrt(m1*m2),现在给定n个这样的生物,求合并成一个的最小重量 思路: m1+m2 >=  2*sqrt(m1*m2) 所以每次取大的去合并,能变小. 直接优先队列就可以啦. #include<cstdio> #include<cmath> #include<queue> using namespace std;…
Stripies Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 10263   Accepted: 4971 Description Our chemical biologists have invented a new very useful form of life called stripies (in fact, they were first called in Russian - polosatiki, bu…
Stripies Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 18198   Accepted: 8175 Description Our chemical biologists have invented a new very useful form of life called stripies (in fact, they were first called in Russian - polosatiki, bu…
Description Our chemical biologists have invented a new very useful form of life called stripies (in fact, they were first called in Russian - polosatiki, but the scientists had to invent an English name to apply for an international patent). The str…
(- ̄▽ ̄)-* #include<iostream> #include<cstdio> #include<cmath> #include<algorithm> using namespace std; int main() { int n; ]; while(~scanf("%d",&n)) { ;i<n;i++) { int m; scanf("%d",&m); w[i]=m*1.0; } s…
每次合并最大的两个,优先级队列维护一下. 输出的时候%.3lf G++会WA,C++能AC,改成%.3f,都能AC. #include<cstdio> #include<cstring> #include<cmath> #include<queue> #include<algorithm> using namespace std; int n,s; priority_queue<double>Q; int main() { while…
题意:科学家发现一种奇怪的东西,他们有重量weight,如果他们碰在一起,总重变成2*sqrt(m1*m2).要求出最终的重量的最小值. 思路:每次选取质量m最大的两个stripy进行碰撞结合,能够得到最小的质量.所有只要维护一个优先队列就可以了 #include <iostream> #include <cstdio> #include <queue> #include <math.h> #include <cstring> #include…
题目链接: PKU:http://poj.org/problem?id=1862 ZJU:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=543 Description Our chemical biologists have invented a new very useful form of life called stripies (in fact, they were first called in Russian -…
题目描述:http://poj.org/problem?id=1862 题目大意:你有n个数要合并,每两个数x,y合并后得到2*sqrt(x*y).求最后留下的一个数的最小值. 每合并一次,就会有数被开方,那么你越早合并的数被开放的次数越多,于是每次把最大的两个数合并即可.用到优先队列. 代码: #include<cstdio> #include<queue> #include<cmath> using namespace std; priority_queue<…
Stripies Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 14151   Accepted: 6628 Description Our chemical biologists have invented a new very useful form of life called stripies (in fact, they were first called in Russian - polosatiki, bu…
Stripies 直接上中文了 Descriptions 我们的化学生物学家发明了一种新的叫stripies非常神奇的生命.该stripies是透明的无定形变形虫似的生物,生活在果冻状的营养培养基平板菌落.大部分的时间stripies在移动.当他们两个碰撞,会有新stripie生成,而旧的不见了.经过长期研究,他们发现新stripies的体重不等于消失的stripies的体重,而是:如果一个质量为m1和m2的stripies相撞,生成的stripies体重是2*sqrt(m1*m2) 现在,科学…
#include <iostream> #include <algorithm> #include <iomanip> #include <cmath> #define MAXN 100 using namespace std; double _m[MAXN]; double fun(); int n; int main() { //freopen("acm.acm","r",stdin); int i; cin>…
题意: 有n个数,要把其中2个数进行2*根号(n1*n2)操作,求剩下最小的那个数是多少? 哭诉:看题目根本没看出来要让我做这个操作. 思路: 每次把最大的,次大的拿出来进行操作 用"优先队列"巧解,优先队列中剩下的那个就是题目要求求的答案. 解题代码: #include <iostream> #include <math.h> #include <algorithm> #include <queue> #include <cstd…
POJ 1862 Stripies https://vjudge.net/problem/POJ-1862 题目:     Our chemical biologists have invented a new very useful form of life called stripies (in fact, they were first called in Russian - polosatiki, but the scientists had to invent an English n…
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Stripies Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 20456   Accepted: 9098 Description Our chemical biologists have invented a new very useful form of life called stripies (in fact, they were first called in Russian - polosatiki, bu…
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Log 2016-3-21 网上找的POJ分类,来源已经不清楚了.百度能百度到一大把.贴一份在博客上,鞭策自己刷题,不能偷懒!! 初期: 一.基本算法: (1)枚举. (poj1753,poj2965) (2)贪心(poj1328,poj2109,poj2586) (3)递归和分治法. (4)递推. (5)构造法.(poj3295) (6)模拟法.(poj1068,poj2632,poj1573,poj2993,poj2996) 二.图算法: (1)图的深度优先遍历和广度优先遍历. (2)最短路…
初期:一.基本算法:     (1)枚举. (poj1753,poj2965)     (2)贪心(poj1328,poj2109,poj2586)     (3)递归和分治法.     (4)递推.     (5)构造法.(poj3295)     (6)模拟法.(poj1068,poj2632,poj1573,poj2993,poj2996)二.图算法:     (1)图的深度优先遍历和广度优先遍历.     (2)最短路径算法(dijkstra,bellman-ford,floyd,hea…
初期: 一.基本算法:      (1)枚举. (poj1753,poj2965)      (2)贪心(poj1328,poj2109,poj2586)      (3)递归和分治法.      (4)递推.      (5)构造法.(poj3295)      (6)模拟法.(poj1068,poj2632,poj1573,poj2993,poj2996) 二.图算法:      (1)图的深度优先遍历和广度优先遍历.      (2)最短路径算法(dijkstra,bellman-ford…
poj 题目分类 按照ac的代码长度分类(主要参考最短代码和自己写的代码) 短代码:0.01K--0.50K:中短代码:0.51K--1.00K:中等代码量:1.01K--2.00K:长代码:2.01K以上. 短:1147.1163.1922.2211.2215.2229.2232.2234.2242.2245.2262.2301.2309.2313.2334.2346.2348.2350.2352.2381.2405.2406: 中短:1014.1281.1618.1928.1961.2054…
本文来自:http://www.cppblog.com/snowshine09/archive/2011/08/02/152272.spx 多版本的POJ分类 流传最广的一种分类: 初期: 一.基本算法: (1)枚举. (poj1753,poj2965) (2)贪心(poj1328,poj2109,poj2586) (3)递归和分治法. (4)递推. (5)构造法.(poj3295) (6)模拟法.(poj1068,poj2632,poj1573,poj2993,poj2996) 二.图算法:…
初期:一.基本算法:     (1)枚举. (poj1753,poj2965)     (2)贪心(poj1328,poj2109,poj2586)     (3)递归和分治法.     (4)递推.     (5)构造法.(poj3295)     (6)模拟法.(poj1068,poj2632,poj1573,poj2993,poj2996)二.图算法:     (1)图的深度优先遍历和广度优先遍历.     (2)最短路径算法(dijkstra,bellman-ford,floyd,hea…
http://www.cnblogs.com/kuangbin/archive/2011/07/29/2120667.html 初期:一.基本算法:     (1)枚举. (poj1753,poj2965)     (2)贪心(poj1328,poj2109,poj2586)     (3)递归和分治法.     (4)递推.     (5)构造法.(poj3295)     (6)模拟法.(poj1068,poj2632,poj1573,poj2993,poj2996)二.图算法:     (…
Hint:补补基础... 初期:一.基本算法:     (1)枚举. (poj1753,poj2965)     (2)贪心(poj1328,poj2109,poj2586)     (3)递归和分治法.     (4)递推.     (5)构造法.(poj3295)     (6)模拟法.(poj1068,poj2632,poj1573,poj2993,poj2996)二.图算法:     (1)图的深度优先遍历和广度优先遍历.     (2)最短路径算法(dijkstra,bellman-f…
初期: 一.基本算法:      (1)枚举. (poj1753,poj2965)      (2)贪心(poj1328,poj2109,poj2586)      (3)递归和分治法.      (4)递推.      (5)构造法.(poj3295)      (6)模拟法.(poj1068,poj2632,poj1573,poj2993,poj2996) 二.图算法:      (1)图的深度优先遍历和广度优先遍历.      (2)最短路径算法(dijkstra,bellman-ford…
以下是poj百道水题,新手可以考虑从这里刷起 搜索1002 Fire Net1004 Anagrams by Stack1005 Jugs1008 Gnome Tetravex1091 Knight Moves1101 Gamblers1204 Additive equations 1221 Risk1230 Legendary Pokemon1249 Pushing Boxes 1364 Machine Schedule1368 BOAT1406 Jungle Roads1411 Annive…
原文地址:北大POJ题库使用指南 北大ACM题分类主流算法: 1.搜索 //回溯 2.DP(动态规划)//记忆化搜索 3.贪心 4.图论 //最短路径.最小生成树.网络流 5.数论 //组合数学(排列组合).递推关系.质因数法 6.计算几何 //凸壳.同等安置矩形的并的面积与周长.凸包计算问题 8.模拟 9.数据结构 //并查集.堆.树形结构 10.博弈论 11.CD有正气法题目分类: 1. 排序 1423, 1694, 1723, 1727, 1763, 1788, 1828, 1838, 1…