Period Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2258    Accepted Submission(s): 1117 Problem Description For each prefix of a given string S with N characters (each character has an ASCII…

上次因为haipz组织的比赛中有道题必须用到KMP算法,因此赛后便了解了下它,在仔细拜读了孤~影神牛的文章之后有种茅塞顿开的感觉,再次ORZ. 附上链接http://www.cnblogs.com/yjiyjige/p/3263858.html. 对于整个KMP算法,最精髓的部分便是关于next数组的生成.一开始ruijia liu的书上贴上的代码感觉完全不能理解,但是看神犇的分析觉得似乎明白了什么故写下此文来理理思路. 在整理思路之前首先记住一句话:next[j]表示的是当在str2[j]!=…

Manacher算法 Manacher模板题…… //HDOJ 3068 #include<cstdio> #include<cstring> #include<cstdlib> #include<iostream> #include<algorithm> #define rep(i,n) for(int i=0;i<n;++i) #define F(i,j,n) for(int i=j;i<=n;++i) #define D(i,j…

FatMouse' Trade Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 56784    Accepted Submission(s): 19009 Problem Description FatMouse prepared M pounds of cat food, ready to trade with the cats g…

Nasty Hacks Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3049    Accepted Submission(s): 2364 Problem Description You are the CEO of Nasty Hacks Inc., a company that creates small pieces of…

Box of Bricks Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5994    Accepted Submission(s): 2599 Problem Description Little Bob likes playing with his box of bricks. He puts the bricks one up…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1358 思路:Next数组的用法,在第i个位置上如果有i%(i-Next[i])==0的话最小循环节就是T[0~i],共有i/(i-Next[i])个循环节 题意就是让从第二个位置开始找出有循环的位置节并输出循环节个数 #include<cstdio> #include<iostream> #include<algorithm> #include<math.h>…

Problem Description Contest time again! How excited it is to see balloons floating around. But to tell you a secret, the judges' favorite time is guessing the most popular problem. When the contest is over, they will count the balloons of each color…

卧槽....最近刷的cf上有最短路,本来想拿这题复习一下.... 题意就是在输出最短路的情况下,经过每个节点会增加税收,另外要字典序输出,注意a到b和b到a的权值不同 然后就是处理字典序的问题,当松弛时发现相同值的时候,判断两条路径的字典序 代码 #include "stdio.h" const int MAXN=110; const int INF=10000000; bool vis[MAXN]; int pre[MAXN]; int cost[MAXN][MAXN],lowcos…

Rectangles    HDOJ(2056) http://acm.hdu.edu.cn/showproblem.php?pid=2056 题目描述:给2条线段,分别构成2个矩形,求2个矩形相交面积. 算法:先用快速排斥判断2个矩形是否相交.若不相交,面积为0.若相交,将x坐标排序去中间2个值之差,y坐标也一样.最后将2个差相乘得到最后结果. 这题是我大一的时候做过的,当时一看觉得很水,写起来发现其实没我想的那么水.分了好几类情况没做出来.今天看了点关于判断线段相交的知识,想起了这题便拿来练…