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
- Linux常见命令整理(一)
整理一下,以备后用 cd /home 进入/home文件夹 cd .. 返回上一级文件夹 cd ../.. 返回上两级文件夹 cd 进入个人的主文件夹 cd - 返回上次所在的文件夹 pwd 显 ...
- 采用Java语言如何实现高速文件复制?
今天review代码也看到了"大神"用老方法来实现文件拷贝.今天归结一下使用Java语言怎样实现高速文件复制: 代码1--使用文件通道的方式: import java.io.Fil ...
- 游戏 TRAP(SNRS)AlphaBeta版本
大家好,我是PuzzledBoy,大一(大二快).我是一个独立的游戏开发商,我的梦想是成为一名伟大的艺术家的第九 今天来公布我的第一个独立游戏TRAP(SNRS)的Alpha測试版啦啦啦~~~! 游戏 ...
- Multicast on Openstack
I test multicast on openstack. I use external Router in this test. Openstack Environment: Havana (ML ...
- 【C语言探索之旅】 第三部分第一课:SDL开发游戏之安装SDL
内容简介 1.课程大纲 2.第三部分第一课: SDL开发游戏之安装SDL 3.第三部分第二课预告: SDL开发游戏之创建窗口和画布 课程大纲 我们的课程分为四大部分,每一个部分结束后都会有练习题,并会 ...
- Atitit.Hibernate于Criteria 使用汇总and 关系查询 and 按照子对象查询 o9o
Atitit.Hibernate于Criteria 使用总结and 关联查询 and 依照子对象查询 o9o 1. Criteria,,Criterion ,, 1 <2. 基本的对象黑头配置磊 ...
- WORD中怎样自己主动生成文件夹?
步骤: 1.输入当做标题的文字 2.将文字设置为标题样式 3.光标放在要加入�文件夹的位置 4.选择插入->引用->索引和文件夹->文件夹->确定
- 合理设置MTU,提升下载速度
可能很少有雷友注意过“本机.网络”的“MTU”值对自己网络性能产生的影响.对于追求更快的下载速度来说,MTU值设置不当,就仿佛穿着高跟鞋跑步一般. MTU是什么? “MTU=最大传输单元 单位:字节” ...
- NYOJ 372 巧克力的
巧克力 时间限制:4000 ms | 内存限制:65535 KB 难度:2 描写叙述 布欧能够把人变成巧克力吃了来添加他的能量,也有可能降低. 如今布欧变了n*m个巧克力,并把巧克力排成一个n*m ...
- 右键菜单中的好友列表Ajax直接跳转请求到登陆页面
今天,我们正在做正确的菜单.当点击重命名Ajax要求,并且不发送数据的背景,但直接跳到主页. 我百思不得其解,后来我发现在头版的一个问题: <li><a href='#' oncli ...