zoukankan      html  css  js  c++  java
  • POJ3690 Constellations 【KMP】

    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
    ", cas++, ans);
        }
        return 0;
    } 
    


  • 相关阅读:
    1219
    hdu 1280 前m大的数 和 hdu 4223 Dynamic Programming?
    codeforces Lucky Sum
    20121028
    asp.net(c#) 取得网卡mac地址
    代码使用FileUpload控件上传图片并自动生成缩略图、自动生成带文字和图片的水印图
    SQL Server日志文件总结及日志满的处理
    服务器应用程序不可用解决方案集
    如何调用存储过程,有返回值的,有参数的,存储过程中调用存储过程。(MS SQL Server)
    让网页自动刷新
  • 原文地址:https://www.cnblogs.com/mfrbuaa/p/4592515.html
Copyright © 2011-2022 走看看