题目传送：Zipper
思路：设状态dp[i][j]为字符串A前i个字符和B前j个字符能否组成C的前i+j个字符，边界为dp[0][0] = 1能则为true，否则false
AC代码：
#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <cmath>#include <queue>#include <stack>#include <vector>#include <map>#include <set>#include <deque>#include <cctype>#define LL long long#define INF 0x7fffffffusing namespace std;char A[205];char B[205];char C[405];int T;int dp[205][205];int main() {scanf("%d", &T);int cas = 1;while(T –) {scanf("%s %s %s", A + 1, B + 1, C + 1);int lena = strlen(A + 1);int lenb = strlen(B + 1);int lenc = strlen(C + 1);memset(dp, 0, sizeof(dp));dp[0][0] = 1;for(int i = 0; i <= lena; i ++) {for(int j = 0; j <= lenb; j ++) {if(i > 0 && A[i] == C[i+j] && dp[i-1][j] == 1) {dp[i][j] = 1;}if(j > 0 && B[j] == C[i+j] && dp[i][j-1] == 1) {dp[i][j] = 1;}}}if(dp[lena][lenb] == 1) {printf("Data set %d: yes\n", cas ++);}else {printf("Data set %d: no\n", cas ++);}}return 0;}
刚开始的思路是算出A与C的LCS和B与C的LCS，如果满足A和B都是C的子序列的话且所有字母都相同数目一样则输出yes，，但是有漏洞，比如ab, ab, baba,对于A是B或B是A的子串的时候就不对了
用LCS做的代码（贴一下，谨记）：
#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <cmath>#include <queue>#include <stack>#include <vector>#include <map>#include <set>#include <deque>#include <cctype>#define LL long long#define INF 0x7fffffffusing namespace std;int T;char s1[205], s2[205], s3[405];int num1[26], num2[26];int dp[205][405], dp2[205][405];int main() {scanf("%d", &T);int cas = 1;while(T –) {memset(num1, 0, sizeof(num1));memset(num2, 0, sizeof(num2));scanf("%s %s %s", s1 + 1, s2 + 1, s3 + 1);int len1 = strlen(s1 + 1);int len2 = strlen(s2 + 1);int len3 = strlen(s3 + 1);for(int i = 1; i <= len1; i ++) {num1[s1[i] – 'a'] ++;}for(int i = 1; i <= len2; i ++) {num1[s2[i] – 'a'] ++;}for(int i = 1; i <= len3; i ++) {num2[s3[i] – 'a'] ++;}printf("Data set %d: ", cas ++);int flag = 1;for(int i = 0; i < 26; i ++) {if(num1[i] != num2[i]) {flag = 0;break;}}if(flag == 0) {printf("no\n");continue;}memset(dp, 0, sizeof(dp));memset(dp2, 0, sizeof(dp2));for(int i = 1; i <= len1; i ++) {for(int j = 1; j <= len3; j ++) {if(s1[i] == s3[j]) {dp[i][j] = dp[i-1][j-1] + 1;}else {dp[i][j] = max(dp[i-1][j], dp[i][j-1]);}}}for(int i = 1; i <= len2; i ++) {for(int j = 1; j <= len3; j ++) {if(s2[i] == s3[j]) {dp2[i][j] = dp2[i-1][j-1] + 1;}else {dp2[i][j] = max(dp2[i-1][j], dp2[i][j-1]);}}}if(dp[len1][len3] != len1 || dp2[len2][len3] != len2) {flag = 0;}if(flag) {printf("yes\n");}else printf("no\n");} return 0;}
当一个人真正觉悟的一刻，他放弃追寻外在世界的财富，
