Constellations
Time Limit: 3000MS   Memory Limit: 65536K
Total Submissions: 5044   Accepted: 983

Description

The starry sky in the summer night is one of the most beautiful things on this planet. People imagine that some groups of stars in the sky form so-called constellations. Formally a constellation is a group of stars that are connected together to form a figure
or picture. Some well-known constellations contain striking and familiar patterns of bright stars. Examples are Orion (containing a figure of a hunter), Leo (containing bright stars outlining the form of a lion), Scorpius (a scorpion), and Crux (a cross).

In this problem, you are to find occurrences of given constellations in a starry sky. For the sake of simplicity, the starry sky is given as a N × M matrix, each cell of which is a '*' or '0' indicating a star in the corresponding position
or no star, respectively. Several constellations are given as a group of T P × Q matrices. You are to report how many constellations appear in the starry sky.

Note that a constellation appears in the sky if and only the corresponding P × Q matrix exactly matches some P × Q sub-matrix in the N × M matrix.

Input

The input consists of multiple test cases. Each test case starts with a line containing five integers N, M, T, P and Q(1 ≤ N, M ≤ 1000, 1 ≤ T ≤ 100, 1 ≤ P, Q ≤ 50). 

The following N lines describe the N × M matrix, each of which contains M characters '*' or '0'.

The last part of the test case describe T constellations, each of which takes P lines in the same format as the matrix describing the sky. There is a blank line preceding each constellation.

The last test case is followed by a line containing five zeros.

Output

For each test case, print a line containing the test case number( beginning with 1) followed by the number of constellations appearing in the sky.

Sample Input

3 3 2 2 2
*00
0**
*00 **
00 *0
**
3 3 2 2 2
*00
0**
*00 **
00 *0
0*
0 0 0 0 0

Sample Output

Case 1: 1
Case 2: 2

题意:给定一个n行m列的01矩阵。再给定t个p行q列的小01矩阵,求这t个小矩阵有多少个在大矩阵中。

题解:这题我用的是KMP,先把矩阵二进制压缩成整型数组,再求整型数组的next数组,再去跟压缩后的大矩阵匹配。遗憾的是TLE了。

这题先就这样放着,等以后学了AC自己主动机再试试。

#include <stdio.h>
#define maxn 1002
#define maxm 52 char bigMap[maxn][maxn], smallMap[maxm][maxm];
__int64 smallToInt[maxm], hash[maxn][maxn];
int m, n, t, p, q, next[maxm]; void toInt64(int i, int j)
{
__int64 sum = 0;
for(int k = 0; k < p; ++k)
if(bigMap[i + k][j] == '*') sum = sum << 1 | 1;
else sum <<= 1;
hash[i][j] = sum;
} void charToHash()
{
int i, j, temp = n - p;
for(i = 0; i <= temp; ++i){
for(j = 0; j < m; ++j) toInt64(i, j);
}
} void getNext()
{
__int64 sum;
int i, j;
for(i = 0; i < q; ++i){
for(sum = j = 0; j < p; ++j)
if(smallMap[j][i] == '*') sum = sum << 1 | 1;
else sum <<= 1;
smallToInt[i] = sum;
}
i = 0; j = -1;
next[0] = -1;
while(i < q){
if(j == -1 || smallToInt[i] == smallToInt[j]){
++i; ++j;
if(smallToInt[i] == smallToInt[j]) next[i] = next[j];
else next[i] = j; //mode 2
}else j = next[j];
}
} bool KMP()
{
getNext();
int i, j, k, temp = n - p;
for(k = 0; k <= temp; ++k){
i = j = 0;
while(i < m && j < q){
if(j == -1 || hash[k][i] == smallToInt[j]){
++i; ++j;
}else j = next[j];
}
if(j == q) return true;
}
return false;
} int main()
{
// freopen("stdin.txt", "r", stdin);
int i, j, ans, cas = 1;
while(scanf("%d%d%d%d%d", &n, &m, &t, &p, &q) != EOF){
if(m + n + t + p + q == 0) break;
for(i = 0; i < n; ++i)
scanf("%s", bigMap[i]);
charToHash(); ans = 0;
while(t--){
for(i = 0; i < p; ++i)
scanf("%s", smallMap[i]);
if(KMP()) ++ans;
}
printf("Case %d: %d\n", cas++, ans);
}
return 0;
}

POJ3690 Constellations 【KMP】的更多相关文章

  1. 【KMP】【最小表示法】NCPC 2014 H clock pictures

    题目链接: http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1794 题目大意: 两个无刻度的钟面,每个上面有N根针(N<=200000),每个 ...

  2. 【动态规划】【KMP】HDU 5763 Another Meaning

    题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5763 题目大意: T组数据,给两个字符串s1,s2(len<=100000),s2可以被解读成 ...

  3. HDOJ 2203 亲和串 【KMP】

    HDOJ 2203 亲和串 [KMP] Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) ...

  4. 【KMP】Censoring

    [KMP]Censoring 题目描述 Farmer John has purchased a subscription to Good Hooveskeeping magazine for his ...

  5. 【KMP】OKR-Periods of Words

    [KMP]OKR-Periods of Words 题目描述 串是有限个小写字符的序列,特别的,一个空序列也可以是一个串.一个串P是串A的前缀,当且仅当存在串B,使得A=PB.如果P≠A并且P不是一个 ...

  6. 【KMP】Radio Transmission

    问题 L: [KMP]Radio Transmission 题目描述 给你一个字符串,它是由某个字符串不断自我连接形成的.但是这个字符串是不确定的,现在只想知道它的最短长度是多少. 输入 第一行给出字 ...

  7. 【kmp】似乎在梦中见过的样子

    参考博客: BZOJ 3620: 似乎在梦中见过的样子 [KMP]似乎在梦中见过的样子 题目描述 「Madoka,不要相信QB!」伴随着Homura的失望地喊叫,Madoka与QB签订了契约. 这是M ...

  8. 【POJ2752】【KMP】Seek the Name, Seek the Fame

    Description The little cat is so famous, that many couples tramp over hill and dale to Byteland, and ...

  9. 【POJ2406】【KMP】Power Strings

    Description Given two strings a and b we define a*b to be their concatenation. For example, if a = & ...

随机推荐

  1. [ACM] hdu 5045 Contest (减少国家Dp)

    Contest Problem Description In the ACM International Collegiate Programming Contest, each team consi ...

  2. UVA 11388-GCD LCM(数学)

    I I U C   O N L I N E   C  Problem D: GCD LCM Input: standard input Output: standard output The GCD ...

  3. Oracle基于学习3--Oracle创建用户和授权

    Oracleserver端的操作,如以下一般: 1)       安装Oracleserver软件 2)       创建数据库(安装时自己主动创建) 3)       配置监听(安装时自己主动配置) ...

  4. mybatis至mysql插入一个逗号包含值误差

    mybatis至mysql插入形如"11,22,33"当误差.我使用了错误的原因是美元符号镶嵌sql.正确的做法是使用#  有时间去看看mybatis的$和#差异. 版权声明:本文 ...

  5. MVC下判断用户登录和授权状态方法

    MVC下判断用户登录和授权状态方法 在我们日常开发的绝大多数系统中,都涉及到管理用户的登录和授权问题.登录功能(Authentication),针对于所有用户都开放:而授权(Authorization ...

  6. Nuget的使用

    前言 最近看到园子上有关于NuGet (读音:new get)的使用,所以心血来潮也跟着学习做了一下,觉得很流b哦.于是也就记一下自己的学习心得(并非一味的重复轮子,只是觉得他人做的写的就是对的,自己 ...

  7. 谈话节目APE系列:如何成为技术达人

    作为一个程序猿,总有消退的前辈.或更年轻的同行.牛逼的人总是羡慕. 让我们搞自己痛苦的日子 BUG .头发很快结束了抓,人们扫两.修改一行代码.问题得以克服:例如,他们自己开发的十年,少付 10K , ...

  8. hive的udf制剂

    1.配置环境 1.介绍jar包裹:hive-exec-0.8.0.jar 和 hadoop-core-0.20.2.jar 2.书写udf性能 1.extends UDF 2.又一次定义UDF的eva ...

  9. Akka.net开发第一个分布式应用

    Akka.net开发第一个分布式应用 系列主题:基于消息的软件架构模型演变 既然这个系列的主题是”基于消息的架构模型演变“,少不了说说Actor模型.Akka.net是一个基于Actor模型的分布式框 ...

  10. hud 1312 Red and Black

    题目: 链接:pid=1312">点击打开链接 题意: DFS搜索 算法: dfs 思路: 简单题 代码: #include<iostream> #include<c ...