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  • 18.06.30 POJ 百练2192:Zipper

    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] == '' && w2[ninw2] == ''&&res[ninres]=='')
    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
    ", kase);
    41     else if(flag==1)
    42         printf("Data set %d: yes
    ", 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
    注定失败的战争,也要拼尽全力去打赢它; 就算输,也要输得足够漂亮。
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  • 原文地址:https://www.cnblogs.com/yalphait/p/9247229.html
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