HDU 1501 Zipper(DP，DFS)
意甲冠军 是否可以由串来推断a,b字符不改变其相对为了获取字符串的组合c
本题有两种解法 DP或者DFS
考虑DP 令d[i][j]表示是否能有a的前i个字符和b的前j个字符组合得到c的前i+j个字符 值为0或者1 那么有d[i][j]=(d[i-1][j]&&a[i]==c[i+j])||(d[i][j-1]&&b[i]==c[i+j]) a,b的下标都是从1開始的 注意0的初始化
#include<cstdio>
#include<cstring>
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;
}
以下是dfs的代码 看是否能在ab中相应搜到c的每个字母就可
//DFS版
#include <cstdio>
#include <cstring>
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".
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.
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.
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
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