96 lines
2.3 KiB
C
96 lines
2.3 KiB
C
#include <stdio.h>
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#include <string.h>
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// 打印LCS的函数
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void PrintLCS(int path[][100], char *x, int i, int j) {
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if (i == 0 || j == 0) {
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return;
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}
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if (path[i][j] == 0) {
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PrintLCS(path, x, i - 1, j - 1);
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printf("%c", x[i - 1]);
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} else if (path[i][j] == 1) {
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PrintLCS(path, x, i - 1, j);
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} else {
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PrintLCS(path, x, i, j - 1);
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}
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}
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// 找到LCS的长度
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int findLCS(int dp[][100], char *A, int n, char *B, int m) {
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// 初始化dp数组
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for (int i = 0; i <= n; i++) {
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for (int j = 0; j <= m; j++) {
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dp[i][j] = 0; // 数组初始化
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}
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}
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for (int i = 1; i <= n; i++) {
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for (int j = 1; j <= m; j++) { // 执行状态转移方程
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if (A[i - 1] == B[j - 1]) {
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dp[i][j] = dp[i - 1][j - 1] + 1;
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} else {
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dp[i][j] = dp[i - 1][j] > dp[i][j - 1] ? dp[i - 1][j] : dp[i][j - 1];
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}
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}
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}
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return dp[n][m];
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}
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int main() {
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char x[] = "w8xta2px0i1jf4jfn8du8dk1fk2kvchd32";
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char y[] = "i3xsmdxwuv6a2yt520";
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int m = strlen(x);
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int n = strlen(y);
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// 创建二维数组
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int c[101][101]; // 101用于存储m+1和n+1的值
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int path[101][101]; // 路径数组
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// 初始化二维数组
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for (int i = 0; i <= m; i++) {
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c[i][0] = 0;
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}
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for (int i = 0; i <= n; i++) {
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c[0][i] = 0;
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}
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// 实现LCS逻辑
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for (int i = 1; i <= m; i++) {
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for (int j = 1; j <= n; j++) {
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if (x[i - 1] == y[j - 1]) {
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c[i][j] = c[i - 1][j - 1] + 1;
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} else if (c[i - 1][j] >= c[i][j - 1]) {
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c[i][j] = c[i - 1][j];
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path[i][j] = 1; // 来自上方
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} else {
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c[i][j] = c[i][j - 1];
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path[i][j] = -1; // 来自左侧
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}
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}
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}
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// 输出c数组
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printf("c:\n");
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for (int i = 0; i <= m; i++) {
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for (int j = 0; j <= n; j++) {
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printf("%d\t", c[i][j]);
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}
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printf("\n");
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}
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// 输出path数组
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printf("path:\n");
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for (int i = 0; i <= m; i++) {
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for (int j = 0; j <= n; j++) {
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printf("%d\t", path[i][j]);
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}
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printf("\n");
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}
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printf("%s与%s的最长公共子序列为：\n", x, y);
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PrintLCS(path, x, m, n);
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return 0;
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} |