zoukankan      html  css  js  c++  java
  • POJ 1458 Common Subsequence

    Description

    A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = < x1, x2, ..., xm > another sequence Z = < z1, z2, ..., zk > is a subsequence of X if there exists a strictly increasing sequence < i1, i2, ..., ik > of indices of X such that for all j = 1,2,...,k, xij = zj. For example, Z = < a, b, f, c > is a subsequence of X = < a, b, c, f, b, c > with index sequence < 1, 2, 4, 6 >. Given two sequences X and Y the problem is to find the length of the maximum-length common subsequence of X and Y.

    Input

    The program input is from the std input. Each data set in the input contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct.

    Output

    For each set of data the program prints on the standard output the length of the maximum-length common subsequence from the beginning of a separate line.

    Sample Input

    abcfbc         abfcab
    programming    contest 
    abcd           mnp

    Sample Output

    4
    2
    0

    最长公共子串问题:

     1 #include<stdio.h>
     2 #include<string.h>
     3 #define maxn 1010
     4 
     5 int d[2][maxn];
     6 char a[maxn], b[maxn];
     7 int main()
     8 {
     9     while(scanf("%s %s", a, b) == 2){
    10         int la, lb, i, j;
    11         la = strlen(a);
    12         lb = strlen(b);
    13         memset(d, 0, sizeof(d));
    14         for(i = 1; i <= la; i++)
    15             for(j = 1; j <= lb; j++)
    16                 if(a[i-1] == b[j-1])
    17                     d[i%2][j] = d[(i-1)%2][j-1] + 1;
    18                 else
    19                     d[i%2][j] = d[i%2][j-1] > d[(i-1)%2][j] ? d[i%2][j-1] : d[(i-1)%2][j];//取最大值
    20         printf("%d
    ", d[la%2][lb]);
    21     }
    22     return 0;
    23 }
    
    

    另解:

    POJ 1458 最长公共子列 - 紫儿的LG - 囧囧

     1 #include <stdio.h>
     2 #include <string.h>
     3 #define MAXN 1010
     4 
     5 char X[MAXN];
     6 char Y[MAXN];
     7 int dp[MAXN][MAXN];
     8 int fmax(int a, int b)
     9 {
    10     return a > b ? a : b;
    11 }
    12 
    13 void dpf(int m,int n)
    14 {
    15     int i, j;
    16     for (i = 1; i <= m; i++)
    17         dp[i][0] = 0;
    18     for (i = 1; i <= n; i++)
    19         dp[0][i] = 0;
    20     for (i = 1; i <= m; i++)
    21         for (j = 1; j <= n; j++)
    22             if ( X[i] == Y[j])
    23                 dp[i][j] = dp[i-1][j-1]+1;
    24             else
    25                 dp[i][j] = fmax(dp[i-1][j], dp[i][j-1]);
    26 }
    27 
    28 int main()
    29 {
    30     int m, n;
    31     while(scanf("%s %s", X, Y) == 2){
    32         m = strlen(X);
    33         n = strlen(Y);
    34     memset(dp,0,sizeof(dp));
    35     dpf(m, n);
    36     printf("%d
    ",dp[m][n]);
    37     }
    38     return 0;
    39 }

    关于最长公共子序列的进一步探讨——输出

     1 #include <stdio.h>
     2 #include <string.h>
     3 
     4 int x[100], y[100], c[100][100], b[100][100];
     5 void LcsLength(int *x,int *y,int m,int n,int c[][100],int b[][100])
     6 {
     7     for(int i = 0; i <= m; i++)
     8         c[i][0] = 0;
     9     for(int j = 0; j <= n; j++)
    10         c[0][j] = 0;
    11     for(int i = 1; i <= m; i++)
    12         for(int j = 1; j <= n; j++){
    13             if(x[i-1] == y[j-1]){
    14                 c[i][j] = c[i-1][j-1]+1;
    15                 b[i][j] = 0;
    16             }
    17             else if(c[i-1][j] >= c[i][j-1]){
    18                 c[i][j] = c[i-1][j];
    19                 b[i][j] = 1;
    20             }
    21             else{
    22                 c[i][j] = c[i][j-1];
    23                 b[i][j] = -1;
    24             }
    25         }
    26 }
    27 
    28 void PrintLCS(int b[][100], int *x, int i, int j)
    29 {
    30     if(i == 0 || j == 0)
    31         return;
    32     if(b[i][j] == 0){
    33         PrintLCS(b, x, i-1, j-1);
    34         printf("%d ", x[i-1]);
    35     }
    36     else if(b[i][j] == 1)
    37         PrintLCS(b, x, i-1, j);
    38     else
    39         PrintLCS(b, x, i, j-1);
    40 }
    41 
    42 int main()
    43 {
    44     int m, n;
    45     scanf("%d", &m);
    46     for(int i = 0; i < m; i++)
    47         scanf("%d", &x[i]);
    48     scanf("%d", &n);
    49     for(int i = 0; i < n; i++)
    50         scanf("%d", &y[i]);
    51     LcsLength(x, y, m, n, c, b);
    52     printf("最长子序列为:");
    53     PrintLCS(b, x, m, n);
    54     printf("
    ");
    55     printf("最长子序列长度为:%d
    ", c[m][n]);
    56     return 0;
    57 }
  • 相关阅读:
    JavaScript设计模式-21.命令模式
    JavaScript设计模式-20.责任链模式
    JavaScript设计模式-18.享元模式
    JavaScript设计模式-19.代理模式
    JavaScript设计模式-17.装饰者模式(下)
    JavaScript设计模式-16.装饰者模式(上)
    面向对象之集合ArrayList
    面向对象之继承
    字符串的添加与切割~~~
    面向对象中构造函数的小练习
  • 原文地址:https://www.cnblogs.com/zzy9669/p/3865958.html
Copyright © 2011-2022 走看看