18.06.30 POJ 百练2192:Zipper

标签：order   names   min   alt   lower   ssi   show   fir   ble   

描述

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".
输入

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.
输出

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.
样例输入

3
cat tree tcraete
cat tree catrtee
cat tree cttaree

样例输出

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

来源

Pacific Northwest 2004

 1 #include <cstdio>
 2 #include <string>
 3 #include <memory.h>
 4 #include <algorithm>
 5 #include <stdlib.h>
 6 #include <math.h>
 7 #include <iostream>
 8 #include<queue>
 9 #include <vector>
10 #include <bitset>
11 using namespace std;
12 
13 char w1[205], w2[205], res[405];
14 int visited[205][205];
15 int kase;
16 int flag;
17 
18 void dfs(int ninw1,int ninw2, int ninres) 
19 {
20     if (w1[ninw1] == ‘\0‘ && w2[ninw2] == ‘\0‘&&res[ninres]==‘\0‘)
21     {
22         flag = 1;
23         return;
24     }
25     if (w1[ninw1]!=res[ninres]&&w2[ninw2]!=res[ninres])
26         return;
27     if (visited[ninw1][ninw2])
28         return;
29     visited[ninw1][ninw2] = 1;
30     if (w1[ninw1] == res[ninres])
31         dfs(ninw1 + 1, ninw2, ninres + 1);
32     if (flag)return;
33     if (w2[ninw2] == res[ninres])
34         dfs(ninw1, ninw2 + 1, ninres + 1);
35 }
36 
37 void solve() {
38     dfs(0,0,0);
39     if(flag==0)
40         printf("Data set %d: no\n", kase);
41     else if(flag==1)
42         printf("Data set %d: yes\n", kase);
43 }
44 
45 int main()
46 {
47     int t;
48     scanf("%d", &t);
49     for(kase=1;kase<=t;kase++){
50         memset(visited, 0, sizeof(int) * 205*205);
51         cin >> w1 >> w2 >> res;
52         flag = 0;
53         solve();
54     }
55     return 0;
56 }

View Code

18.06.30 POJ 百练2192:Zipper

标签：order   names   min   alt   lower   ssi   show   fir   ble   

原文地址：https://www.cnblogs.com/yalphait/p/9247229.html