HDU 1501 Zipper(DP，DFS)

題意 判斷能否由字符串a,b中的字符不改變各自的相對順序組合得到字符串c

本題有兩種解法 DP或者DFS

考慮DP 令d[i][j]表示能否有a的前i個字符和b的前j個字符組合得到c的前i+j個字符 值為0或者1 那麼有d[i][j]=(d[i-1][j]&&a[i]==c[i+j])||(d[i][j-1]&&b[i]==c[i+j]) a,b的下標都是從1開始的 注意0的初始化

#include
#include
using namespace std;
const int N = 205;
char a[N], b[N], c[2 * N];
bool  d[N][N];

int main()
{
    int cas;
    scanf ("%d", &cas);
    for (int k = 1; k <= cas; ++k)
    {
        scanf ("%s%s%s", a + 1, b + 1, c + 1);
        int la = strlen (a + 1), lb = strlen (b + 1), i = 1, j = 1;
        memset (d, 0, sizeof (d));

        while (a[i] == c[i] && i <= la)
            d[i++][0] = true;
        while (b[j] == c[j] && j <= lb)
            d[0][j++] = true;
        for (int i = 1; i <= la; ++i)
            for (int j = 1; j <= lb; ++j)
                d[i][j] = ( (d[i - 1][j] && a[i] == c[i + j]) || (d[i][j - 1] && b[j] == c[i + j]));

        printf ("Data set %d: ", k);
        printf (d[la][lb] ? "yes\n" : "no\n");
    }
    return 0;
}

下面是dfs的代碼 看能否在ab中對應搜到c的每一個字母就可

//DFS版
#include 
#include 
using namespace std;
const int N = 205;
char a[N], b[N], c[2 * N];
bool vis[N][N], ans;
void dfs (int i, int j, int k)
{
    if (c[k] == '\0') ans = true;
    if (ans || vis[i][j])  return ;
    vis[i][j] = true;
    if (a[i] == c[k]) dfs (i + 1, j, k + 1);
    if (b[j] == c[k]) dfs (i, j + 1, k + 1);
}
int main()
{
    int cas;
    scanf ("%d", &cas);
    for (int ca = 1; ca <= cas; ++ca)
    {
        ans = false;
        memset (vis, 0, sizeof (vis));
        scanf ("%s%s%s", a, b, c);
        dfs (0, 0, 0);
        printf ("Data set %d: ", ca);
        printf (ans ? "yes\n" : "no\n");
    }
    return 0;
}

Zipper

Problem Description 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".

Input 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.


Output 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.

Sample Input

3
cat tree tcraete
cat tree catrtee
cat tree cttaree

Sample Output

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