POJ3690 Constellations 【KMP】
| 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
- 【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 ...
- 【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 ...
- 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) ...
- 【KMP】Censoring
[KMP]Censoring 题目描述 Farmer John has purchased a subscription to Good Hooveskeeping magazine for his ...
- 【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 ...
- 【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 ...
- 【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 ...
- 【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 ...
- 【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
- 自动同步Android源代码的脚本(repo sync)
#!/bin/bash echo "================start repo sync====================" repo sync -j5 ]; do ...
- 乐在其中设计模式(C#) - 建造者模式(Builder Pattern)
原文:乐在其中设计模式(C#) - 建造者模式(Builder Pattern) [索引页][源码下载] 乐在其中设计模式(C#) - 建造者模式(Builder Pattern) 作者:webabc ...
- 直接选择排序----java实现
直接选择排序思路: 从待排序数据中选择第一个假定为最小的下标,然后他后面的与他循环比较,得到真的最小值下标,然后最小值前的那一区段依次后移,并把最小值赋值给第一个元素.第二次时,假定第二个为最小,然后 ...
- 数据结构 - AVL木
在计算机科学,AVL木是一个平衡树最早发明. 于AVL树节点,而不管是什么的两个子树之一的高度之间最大的区别,因此,它也被称为平衡树高.查找.O(log n). 插入和移除可能需要一个或更多次通过旋转 ...
- java设计模式演示示例
创建一个模式 1.工厂方法模式(Factory Method) 该程序创建的操作对象,独自一人走出流程,创建产品工厂接口.实际的工作转移到详细的子类.大大提高了系统扩展的柔性,接口的抽象化处理给相互 ...
- Poj3414广泛搜索
<span style="color:#330099;">/* D - D Time Limit:1000MS Memory Limit:65536KB 64bit I ...
- DisplayContent、StackBox、TaskStack笔记
文章仅零散记录自己的一点理解,仅供自己參考. 每一个显示设备,都有一个Display对象,DisplayManagerService专门管理这些Display. 1.DisplayContent() ...
- [原创].NET 业务框架开发实战之六 DAL的重构
原文:[原创].NET 业务框架开发实战之六 DAL的重构 .NET 业务框架开发实战之六 DAL的重构 前言:其实这个系列还是之前的".NET 分布式架构开发实战 ",之所以改了 ...
- http协议和web本质(转)
当你在浏览器地址栏敲入“http://www.cnblogs.com/”,然后猛按回车,呈现在你面前的,将是博客园的首页了(这真是废话,你会认为这是理所当然的).作为一个开发者,尤其是web开发人员, ...
- SplashScreenDemo
对Java应用最常见的抱怨就是启动时间太长.这是因为Java虚拟机花费一段时间去加载所有必需的类,特别是对Swing应用,它们需要从Swing和AWT类库代码中去抽取大量的内容. 用户并不喜欢应用程序 ...