hdoj1501Zipper【dfs】
2015-12-28 15:37
627 查看
Zipper
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8489 Accepted Submission(s): 2997
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
#include<cstdio>
#include<cstdlib>
#include<cstring>
using namespace std;
bool vis[210][210];
char a[210],b[210],c[410];
int la,lb,lc;
bool dfs(int i,int j,int k){
if(k==lc){
return true;
}
if(vis[i][j])return false;
vis[i][j]=true;
if(i<la){
if(a[i]==c[k]&&dfs(i+1,j,k+1))return true;
}
if(j<lb){
if(b[j]==c[k]&&dfs(i,j+1,k+1))return true;
}
return false;
}
int main()
{
int t,i,j,k=1;
scanf("%d",&t);
while(t--){
scanf("%s %s %s",a,b,c);
memset(vis,false,sizeof(vis));
la=strlen(a);
lb=strlen(b);
lc=strlen(c);
printf("Data set %d: ",k++);
if(dfs(0,0,0))
printf("yes\n");
else
printf("no\n");
}
return 0;
}
相关文章推荐
- 软件测试工程师的角度看论证学问
- JMM--(1)基础
- Java对象序列化
- Unity树木生成器
- Android实现Banner界面广告图片循环轮播(包括实现手动滑动循环)
- LeetCode 99:Recover Binary Search Tree
- Fragment初识之静态显示
- 华为机试day3
- 【QT相关】类头文件解读、QT编辑模式、读取text文本
- BroadcastReceiver
- 兄弟连兄弟会
- linux 权限以及权限管理
- Linux 修改只读文件
- LogStash日志分析展示系统
- 修改tomcat默认的编码方式
- 解决Tomcat可以在eclipse启动,却无法显示默认页面
- 深入浅出RxJava(一:基础篇)
- 集成支付宝钱包支付iOS SDK的方法与经验
- WPF Touch操作滚动条,Window弹跳
- images