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动态规划之最长公共子序列算法

技术2022-05-11  0

动态规划之最长公共子序列算法

算法思想: 假设X = (x1, x2, ..., xm), Y = {y1, y2, ..., yn)的最长公共子序列为Z = (z1. z2, ..., zk). 则: 1.若xm = yn, 则zk = xm = yn,且Zk-1是Xm-1和Yn-1的最长公共子序列; 2.若xm != yn, 且zk != xm, 则Z是Xm-1和Y的最长公共子序列; 3.若xm != yn, 且zk != yn, 则Z是X和Yn-1的最长公共子序列。 其中:Xm-1 = (x1, x2, ..., xm-1), Yn-1 = (y1, y2, ..., yn-1), Zk-1 = (z1, z2, ..., zk-1). c[i][j]存储的是Xi和Yj的最长公共子序列的长度,b[i][j]存储的是c[i][j]的值是由哪一个子问题(如上3个子问题)的解得到的。

#include <stdio.h>#define MAX 100char x[MAX+1], y[MAX+1];int b[MAX+1][MAX+1], c[MAX+1][MAX+1], m, n;static void init_xy(void);static void lcs_length(void);static void lcs(int i, int j);int main(int argc, char **argv){        init_xy();        lcs_length();        printf("The lowest common subsequence is: ");        lcs(m, n);        printf("/n");        return 0;}static void init_xy(void){        int i;        printf("Please input two sequences numbers: ");        scanf("%d %d", &m ,&n);        getchar();        printf("Please input two sequences:/n");        for(i = 1; i <= m; i++){                scanf("%c", &x[i]);        }        getchar();        for(i = 1; i <= n; i++){                scanf("%c", &y[i]);        }        getchar();}static void lcs_length(void){        int i, j;        for(i = 1; i <= m; i++){                for(j = 1; j <= n; j++){                        if(x[i] == y[j]){                                c[i][j] = c[i-1][j-1] + 1;                                b[i][j] = 1;                        }                        else if(c[i-1][j] >= c[i][j-1]){                                c[i][j] = c[i-1][j];                                b[i][j] = 2;                        }                        else{                                c[i][j] = c[i][j-1];                                b[i][j] = 3;                        }                }        }        printf("The lowest common subsequence lenght = %d/n", c[m][n]);}static void lcs(int i, int j){        if(i == 0 || j == 0){                return;        }        if(b[i][j] == 1){                lcs(i-1, j-1);                printf("%c ", x[i]);        }        else if(b[i][j] == 2){                lcs(i-1, j);        }        else{                lcs(i, j-1);        }}

执行结果:

[xxxx@localhost suanfa]$ ./a.outPlease input two sequences numbers: 7 6Please input two sequences:ABCBDABBDCABAThe lowest common subsequence lenght = 4The lowest common subsequence is: B C B A


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