1010. Zipper

TAG 动态规划 dp

使3个字符串A B C

dp[i][j]表示A的前i个字符和B的前j个字符能否组合成C的前i+j个字符

状态转移方程：

dp[i][j]= dp[i-1][j] && C[i+j-1]==A[i-1] || dp[i][j-1] && C[i+j-1]==B[j-1]; //注意边界

/* source code of submission 419888, Zhongshan University Online Judge System */ #include <stdio.h> #include <memory.h> #include <string.h> bool dp[201][201]; char A[201],B[201],C[401]; int T,n; int len; int main(int argc, char *argv[]) { scanf("%d", &T); n=0; while ( n<T ) { scanf("%s%s%s", A, B, C); memset(dp, false, sizeof(dp) ); dp[0][0]=true; len=strlen(A); for (int i=0; i<=len; ++i) { for (int j=0; j<=len; ++j) { if ( i+j==0 ) { continue; } if ( i==0 ) { dp[0][j]=dp[0][j-1] && C[j-1]==B[j-1]; } else if ( j==0 ) { dp[i][0]=dp[i-1][0] && C[i-1]==A[i-1]; } else { dp[i][j]= dp[i-1][j] && C[i+j-1]==A[i-1] || dp[i][j-1] && C[i+j-1]==B[j-1]; } //printf("%d", dp[i][j]); } //printf("/n"); } printf("Data set %d: %s/n",++n, dp[len][len]? "yes": "no" ); } return 0;