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. linux kernel 结构体赋值方法{转载}

    原文地址: http://www.chineselinuxuniversity.net/articles/48226.shtml 这几天看Linux的内核源码,突然看到init_pid_ns这个结构体 ...

  2. 一道看似简单的sql需求(转)

    听说这题难住大批高手,你也来试下吧.ps:博问里的博友提出的. 原始数据 select * from t_jeff t  简单排序后数据 select * from t_jeff t order by ...

  3. android存储阵列数据SharedPreferences

    假设要数组数据(如boolean[] .int[]等)到SharedPreferences时,我们能够先将数组数据组织成json数据存储到SharedPreferences,读取时则对json数据进行 ...

  4. [渣译文] SignalR 2.0 系列: 支持的平台

    原文:[渣译文] SignalR 2.0 系列: 支持的平台 英文渣水平,大伙凑合着看吧,并不是逐字翻译的…… 这是微软官方SignalR 2.0教程Getting Started with ASP. ...

  5. chrome(转)

    阅读目录 Chrome的隐身模式 Chrome下各种组合键 Chrome的about指令 chrome://accessibility     查看浏览器当前访问的标签 chrome://appcac ...

  6. resharper 设置代码颜色

  7. struts2移除标签button的id传统的价值观念问题

    <!--显示数据列表--> <tbody id="TableData" class="dataContainer" datakey=" ...

  8. javascript事件和事件处理

    于js期间事件处理被分成三个步骤: 1.发生事件 2.启动事件处理程序 3.事件处理程序做出反应 事件处理程序的调用 1.在javascript中 在javascript中调用事件处理程序,首先要获得 ...

  9. android中怎么把自己须要的app启动图标集中到一个弹出框中

    先看效果图 这个是我们自己的apk点击之后的效果 下边是布局文件 activity_main.xml主布局文件 <LinearLayout xmlns:android="http:// ...

  10. SQL Server 连接问题-命名管道

    原文:SQL Server 连接问题-命名管道 出自:http://blogs.msdn.com/b/apgcdsd/archive/2011/01/12/sql-server-1.aspx 一.前言 ...