POJ 2181】的更多相关文章

#include <iostream> #include <cstdio> #include <cmath> #define MAXN 150005 #include <algorithm> using namespace std; struct node { int num; int sum; bool boo; }; node _m[MAXN]; int n; void init(); int main() { //freopen("acm.a…
Jumping Cows Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6398   Accepted: 3828 Description Farmer John's cows would like to jump over the moon, just like the cows in their favorite nursery rhyme. Unfortunately, cows can not jump. The…
思路: 贪心 对于每个波浪 ans+=最大值-最小值 注意最后一定是选最大值 //By SiriusRen #include <cstdio> using namespace std; int n,a[150500],flag,ans; int main(){ scanf("%d",&n); for(int i=1;i<=n;i++)scanf("%d",&a[i]); for(int i=1;i<=n;i++) if(fla…
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:…
此文转载别人,希望自己能够做完这些题目! 1.POJ动态规划题目列表 容易:1018, 1050, 1083, 1088, 1125, 1143, 1157, 1163, 1178, 1179, 1189, 1208, 1276,1322, 1414, 1456, 1458, 1609, 1644, 1664, 1690, 1699, 1740(博弈),1742, 1887, 1926(马尔科夫矩阵,求平衡), 1936, 1952, 1953, 1958, 1959, 1962, 1975,…
[1]POJ 动态规划题目列表 容易: 1018, 1050, 1083, 1088, 1125, 1143, 1157, 1163, 1178, 1179, 1189, 1208, 1276, 1322, 1414, 1456, 1458, 1609, 1644, 1664, 1690, 1699, 1740(博弈), 1742, 1887,1926(马尔科夫矩阵,求平衡), 1936, 1952, 1953, 1958, 1959, 1962, 1975, 1989, 2018, 2029,…
acm之pku题目分类 对ACM有兴趣的同学们可以看看 DP:  1011   NTA                 简单题  1013   Great Equipment     简单题  1024   Calendar Game       简单题  1027   Human Gene Functions   简单题  1037   Gridland            简单题  1052   Algernon s Noxious Emissions 简单题  1409   Commun…
]POJ 动态规划题目列表 容易: 1018, 1050, 1083, 1088, 1125, 1143, 1157, 1163, 1178, 1179, 1189, 1208, 1276, 1322, 1414, 1456, 1458, 1609, 1644, 1664, 1690, 1699, 1740(博弈), 1742, 1887, 1926(马尔科夫矩阵,求平 衡), 1936,1952, 1953, 1958, 1959, 1962, 1975, 1989, 2018, 2029,2…
此文转载别人,希望自己可以做完这些题目. 1.POJ动态规划题目列表 easy:1018, 1050, 1083, 1088, 1125, 1143, 1157, 1163, 1178, 1179, 1189, 1208, 1276,1322, 1414, 1456, 1458, 1609, 1644, 1664, 1690, 1699, 1740(博弈),1742, 1887, 1926(马尔科夫矩阵,求平衡), 1936, 1952, 1953, 1958, 1959, 1962, 1975…
转载:from: POJ:http://blog.csdn.net/qq_28236309/article/details/47818407 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…
最近想从头开始刷点基础些的题,正好有个网站有关于各大oj的题目分类(http://www.pythontip.com/acm/problemCategory),所以写了点脚本把hdu和poj的一些题目链接按分类爬下来,然后根据题目的AC数目来作为难度指标进行从易到难的排序: POJ       题目标号  通过数 搜索: 1011 336071664 201111321 197841753 185021979 182532386 161761742 122131915 120101579 950…
Halloween treats Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 7644   Accepted: 2798   Special Judge Description Every year there is the same problem at Halloween: Each neighbour is only willing to give a certain total number of sweets…
Find a multiple Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7192   Accepted: 3138   Special Judge Description The input contains N natural (i.e. positive integer) numbers ( N <= 10000 ). Each of that numbers is not greater than 15000…
The Pilots Brothers' refrigerator Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 22286   Accepted: 8603   Special Judge Description The game “The Pilots Brothers: following the stripy elephant” has a quest where a player needs to open a…
Flip Game Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 37427   Accepted: 16288 Description Flip game is played on a rectangular 4x4 field with two-sided pieces placed on each of its 16 squares. One side of each piece is white and the…
Corn Fields Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 9806   Accepted: 5185 Description Farmer John has purchased a lush new rectangular pasture composed of M by N (1 ≤ M ≤ 12; 1 ≤ N ≤ 12) square parcels. He wants to grow some yumm…
Sum of Consecutive Prime Numbers Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 20050   Accepted: 10989 Description Some positive integers can be represented by a sum of one or more consecutive prime numbers. How many such representatio…
Tree Recovery Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 11939   Accepted: 7493 Description Little Valentine liked playing with binary trees very much. Her favorite game was constructing randomly looking binary trees with capital le…
Seek the Name, Seek the Fame Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 17898   Accepted: 9197 Description The little cat is so famous, that many couples tramp over hill and dale to Byteland, and asked the little cat to give names t…
题目: poj 2352 Stars 数星星 题意:已知n个星星的坐标.每个星星都有一个等级,数值等于坐标系内纵坐标和横坐标皆不大于它的星星的个数.星星的坐标按照纵坐标从小到大的顺序给出,纵坐标相同时则按照横坐标从小到大输出. (0 <= x, y <= 32000) 要求输出等级0到n-1之间各等级的星星个数. 分析: 这道题不难想到n平方的算法,即从纵坐标最小的开始搜,每次找它前面横坐标的值比它小的点的个数,两个for循环搞定,但是会超时. 所以需要用一些数据结构去优化,主要是优化找 横坐…
poj   1251  Jungle Roads  (最小生成树) Link: http://poj.org/problem?id=1251 Jungle Roads Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 23507   Accepted: 11012 Description The Head Elder of the tropical island of Lagrishan has a problem. A b…
Kaka's Matrix Travels Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9567   Accepted: 3888 Description On an N × N chessboard with a non-negative number in each grid, Kaka starts his matrix travels with SUM = 0. For each travel, Kaka mo…
Friendship Time Limit: 2000MS   Memory Limit: 20000K Total Submissions: 10626   Accepted: 2949 Description In modern society, each person has his own friends. Since all the people are very busy, they communicate with each other only by phone. You can…
Ikki's Story I - Road Reconstruction Time Limit: 2000MS   Memory Limit: 131072K Total Submissions: 7659   Accepted: 2215 Description Ikki is the king of a small country – Phoenix, Phoenix is so small that there is only one city that is responsible fo…
http://poj.org/problem?id=1144 题意:给你一些点,某些点直接有边,并且是无向边,求有多少个点是割点 割点:就是在图中,去掉一个点,无向图会构成多个子图,这就是割点 Tarjan算法求割点的办法 如果该点为根,那么它的子树必须要大于1 如果该点不为根,那么当low[v]>=dnf[u]时,为割点 Low[v]>=dnf[u]也就是说明U的子孙点只能通过U点访问U的祖先点 #include <stdio.h> #include <stack>…
http://poj.org/problem?id=3614 题意:有n头奶牛想要晒太阳,但他们每个人对太阳都有不同的耐受程度,也就是说,太阳不能太大也不能太小,现在有一种防晒霜,涂抹这个防晒霜可以把太阳的强度固定到一个值 求一共有多少头奶牛可以晒太阳 #include <stdio.h> #include <queue> #include <stdlib.h> using namespace std; int m,n; struct co{ int mi,ma; }c…
链接:http://acm.fzu.edu.cn/problem.php?pid=2181 思路:乍一看以为是并查集,仔细想了下又找不到让函数结束的条件,所以就看了其他人的搜索大法 #include <iostream> #include <stdio.h> #include <stdlib.h> #include <string.h> #include <math.h> #include <queue> #include <a…
POJ 3669 去看流星雨,不料流星掉下来会砸毁上下左右中五个点.每个流星掉下的位置和时间都不同,求能否活命,如果能活命,最短的逃跑时间是多少? 思路:对流星雨排序,然后将地图的每个点的值设为该点最早被炸毁的时间 #include <iostream> #include <algorithm> #include <queue> #include <cstring> using namespace std; #define INDEX_MAX 512 int…
POJ 3009 题意: 给出一个w*h的地图,其中0代表空地,1代表障碍物,2代表起点,3代表终点,每次行动可以走多个方格,每次只能向附近一格不是障碍物的方向行动,直到碰到障碍物才停下来,此时障碍物也会随之消失,如果行动时超出方格的界限或行动次数超过了10则会game over .如果行动时经过3则会win,记下此时行动次数(不是行动的方格数),求最小的行动次数 #include<cstdio> #include<iostream> #include<cstring>…