搜索专题: HDU1501Zipper

最新推荐文章于 2019-04-07 13:47:00 发布

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最新推荐文章于 2019-04-07 13:47:00 发布

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文章标签： java

原文链接：http://www.cnblogs.com/Pretty9/p/7347711.html

本文介绍了一个判断第三字符串是否能由前两个字符串按顺序交替组合而成的问题。通过深度优先搜索算法实现，确保每个字符顺序不变的同时检查所有可能的组合方式。

Zipper

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10465 Accepted Submission(s): 3762

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
  

Source

Pacific Northwest 2004

Problem : 1501 ( Zipper ) Judge Status : Accepted
RunId : 21249294 Language : G++ Author : hnustwanghe
Code Render Status : Rendered By HDOJ G++ Code Render Version 0.01 Beta

#include<iostream> #include<cstring> #include<cstdio> using namespace std; const int N = 200 + 5; char a[N],b[N],c[2*N]; bool visit[N][N],can_find; void DFS(int i,int j,int k){ if(can_find || visit[i][j]) return; if(c[k]=='\0'){can_find = true;return;} visit[i][j] = true; if(a[i]!='\0' && a[i] == c[k]) DFS(i+1,j,k+1); if(b[j]!='\0' && b[j] == c[k]) DFS(i,j+1,k+1); } int main(){ int T,cnt=0; scanf("%d",&T); while(T--){ scanf("%s %s %s",a,b,c); memset(visit,0,sizeof(visit)); can_find = false; DFS(0,0,0); printf("Data set %d: %s\n",++cnt,can_find?"yes":"no"); } return 0; }

#include<iostream>
#include<cstring>
#include<cstdio>

using namespace std;
const int N = 200 + 5;
char a[N],b[N],c[2*N];
bool visit[N][N],can_find;

void DFS(int i,int j,int k){
    if(can_find || visit[i][j]) return;
    if(c[k]=='\0'){can_find = true;return;}
    visit[i][j] = true;
    if(a[i]!='\0' && a[i] == c[k])
        DFS(i+1,j,k+1);
    if(b[j]!='\0' && b[j] == c[k])
        DFS(i,j+1,k+1);
}
int main(){
    int T,cnt=0;
    scanf("%d",&T);
    while(T--){
        scanf("%s %s %s",a,b,c);
        memset(visit,0,sizeof(visit));
        can_find = false;
        DFS(0,0,0);
        printf("Data set %d: %s\n",++cnt,can_find?"yes":"no");
    }
    return 0;
}