HDU 1501 Zipper【DP】

[align=left]Problem Description[/align]
Given three strings, you are to determine whether the third string can be formed by combining the characters in the first two strings. The first two strings can be mixed arbitrarily, but each must stay in its original order.

For example, consider forming "tcraete" from "cat" and
"tree":

String A: cat
String B: tree
String C:
tcraete

As you can see, we can form the third string by alternating
characters from the two strings. As a second example, consider forming "catrtee"
from "cat" and "tree":

String A: cat
String B: tree
String C:
catrtee

Finally, notice that it is impossible to form "cttaree" from
"cat" and "tree".

[align=left]Input[/align]
The first line of input contains a single positive
integer from 1 through 1000. It represents the number of data sets to follow.
The processing for each data set is identical. The data sets appear on the
following lines, one data set per line.

For each data set, the line of
input consists of three strings, separated by a single space. All strings are
composed of upper and lower case letters only. The length of the third string is
always the sum of the lengths of the first two strings. The first two strings
will have lengths between 1 and 200 characters, inclusive.

[align=left]Output[/align]
For each data set, print:

Data set n:
yes

if the third string can be formed from the first two, or

Data
set n: no

if it cannot. Of course n should be replaced by the data set
number. See the sample output below for an example.

[align=left]Sample Input[/align]

3
cat tree tcraete
cat tree catrtee
cat tree cttaree

[align=left]Sample Output[/align]

Data set 1: yes
Data set 2: yes
Data set 3: no

思路

题意是说：判断强两个字符串是否能由第三个拆成。Dp[i][j]的值只有2个，0,（表示串3不能拆成1的i之前2的j之前的串）即判断dp[len(1)][len(2)]是0还是1即可。 dp[i][j]的值可由两个方向而来，即串3的那个字母与串1相等且dp[i-1][j]==1则dp[i][j]==1;串三的字母与串2相等且dp[i][j-1]==1则dp[i][j]==1 （解题报告写的不是很好，忏悔一下）

源码

#include <iostream> #define MAX 210 using namespace std; char a[MAX], b[MAX], c[2*MAX]; int dp[MAX][MAX]; int main() { int i, j, t, cas; scanf("%d", &cas); for(t = 1; t <= cas; t++) { scanf("%s%s%s", a, b, c); memset(dp, 0, sizeof(dp)); dp[0][0] = 1; int la, lb; la = strlen(a); lb = strlen(b); for(i = 0; i <= la; i++) { for(j = 0; j <= lb; j++) { if(i > 0 && a[i-1] == c[i+j-1] && dp[i-1][j]) dp[i][j] = 1; if(j > 0 && b[j-1] == c[i+j-1] && dp[i][j-1]) dp[i][j] = 1; } } printf("Data set %d: ", t); if(dp[la][lb]) printf("yes\n"); else printf("no\n"); } return 0; }