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cocos的事件分发器CCTouchDispatcher，存在两个通道，m_pTargetedHandlers存储CMenu，CScrollView的事件处理器，这里的处理器，在处理过消息后，会声明bClaimed为True，表示事件有被处理过，如果处理器isSwallowsTouches为真，则处理过的消息将被吞噬掉，并且直接跳出对m_pTargetedHandlers的遍历，并且会把消息从消息列表中删除，后续的标准消息处理器将不会收到吞噬的消息。（menu默认会吞噬，CScrollView默认不会） m_pStandardHandlers存储layer的事件处理器，并且，没有吞噬消息的逻辑 阅读全文
posted @ 2013-08-03 22:17
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这题太欢乐了......虽然wa了几次，但是想到骑士在两幅图的传送门中传来传去就觉得这骑士太坑了#include #include #include #include using namespace std;int n,m,cost,head,tail,ans;char map[2][15][15];int sum[2][15][15];struct node { int z,x,y;} q[11111];node st, end;int dirx[4] = {1,-1,0,0};int diry[4] = {0,0,1,-1};void init() { memset(sum,-... 阅读全文
posted @ 2013-08-03 22:15
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Problem A: Football (aka Soccer) The Problem Football the most popular sport in the world (americans insist to call it "Soccer", but we will call it "Football"). As everyone knows, Brasil is the country that have most World Cup titles (four of them: 1958, 1962, 1970 and 1994). As 阅读全文
posted @ 2013-08-03 22:13
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Problem Description The TV shows such as You Are the One has been very popular. In order to meet the need of boys who are still single, TJUT hold the show itself. The show is hold in the Small hall, so it attract a lot of boys and girls. Now there are n boys enrolling in. At the beginning, the n boy 阅读全文
posted @ 2013-08-03 22:11
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The die is castInterGames is a high-tech startup company that specializes in developing technology that allows users to play games over the Internet. A market analysis has alerted them to the fact that games of chance are pretty popular among their potential customers. Be it Monopoly, ludo or backga 阅读全文
posted @ 2013-08-03 22:09
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Problem D: XYZZYADVENT: /ad�vent/, n.The prototypical computer adventure game, first designed by Will Crowther on the PDP-10 in the mid-1970s as an attempt at computer-refereed fantasy gaming, and expanded into a puzzle-oriented game by Don Woods at Stanford in 1976. (Woods had been one of the autho 阅读全文
posted @ 2013-08-03 22:07
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我以为我会是最坚强的那一个 我还是高估了自己 我以为你会是最无情的那一个 还是我贬低了自己 就算不能够在一起 我还是为你担心 就算你可能听不清 也代表我的心意 那北极星的眼泪 闪过你曾经的眼角迷离 那玫瑰花的葬礼 埋葬的却是关于你的回忆 如果时光可以倒流 我希望不要和你分离 如果注定分离 我希望不要和你相遇 ——摘自《小Q失恋日记 》第17卷520页 这是码农小Q第58次失恋了，也是陷得最深的一次。 要知道，小Q自从第一次到腾讯公司报到，就被风姿绰约的前台MM彻底迷住了，这1000多个日日夜夜他无时无刻不在憧憬着他们美好的未来。为了能见到MM，他每天早到晚归，甘愿加班，连续... 阅读全文
posted @ 2013-08-03 22:05
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题目链接： http://acm.hdu.edu.cn/showproblem.php?pid=1885题目意思：给一个矩阵，给一个起点多个终点，有些点有墙不能通过，有些点的位置有门，需要拿到相应颜色的钥匙才能打开，问到达终点的最短步数。解题思路：BFS+状态压缩。将每种颜色对应一个二进制数位，1表示已经得到该颜色的钥匙，0表示没有得到。一把钥匙可以同种颜色的多扇门。代码：#include#include#include#include#include#include#include#include#include#include#include#include#include#define. 阅读全文
posted @ 2013-08-03 22:03
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Problem A: The Monocycle A monocycle is a cycle that runs on one wheel and the one we will be considering is a bit more special. It has a solid wheel colored with five different colors as shown in the figure: The colored segments make equal angles (72o) at the center. A monocy... 阅读全文
posted @ 2013-08-03 22:01
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题意：给你一些括号，问匹配规则成立的括号的个数。思路：这题lrj的黑书上有，不过他求的是添加最少的括号数，是的这些括号的匹配全部成立。我想了下，其实这两个问题是一样的，我们可以先求出括号要匹配的最少数量，那么设原来括号的数量为l , 添加了l' 。那么其实原来括号匹配成功的括号数就是((l + l') / 2 - l') * 2。#define N 105char a[N] ;int dp[N][N] ;int f[N][N] ;int check(int i ,int j) { if(a[i] == '[' && a[j] == 阅读全文
posted @ 2013-08-03 21:59
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