dfs 记忆搜索——注意剪枝方式

今天被一个题磨了一个下午，话不多说先看题

http://acm.hdu.edu.cn/showproblem.php?pid=1501

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不翻译了，比较简单。。。。

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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

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本题用dfs+剪枝就可以。。。。

这是出现TLE的代码

 1 include<iostream>
 2 #include<cstring>
 3 using namespace std;
 4 int vis[201][201];
 5 string str1,str2,str3;
 6 int dfs(int i1,int i2,int p){
 7     if(p==str3.length()){
 8         return 1;
 9     }
10         
11     if(vis[i1][i2]||str1[i1]==str3[p]&&dfs(i1+1,i2,p+1)){
12         vis[i1][i2]=1;
13         return 1;
14     }
15     if(vis[i1][i2]||str2[i2]==str3[p]&&dfs(i1,i2+1,p+1)){
16         vis[i1][i2]=1;
17         return 1;
18     }
19     
20     return 0;
21     
22 }
23 int main(){
24     int T,i=1;
25     cin>>T;
26     while(T--){
27         cin>>str1>>str2>>str3;
28         memset(vis,0,sizeof(vis));
29         
30         if(dfs(0,0,0))
31             cout<<"Data set "<<i<<": yes"<<endl;
32         else
33             cout<<"Data set "<<i<<": no"<<endl;
34         i++;
35     }
36 } 

View Code

这是ac的

 1 #include<iostream>
 2 #include<cstring>
 3 using namespace std;
 4 int vis[201][201];
 5 string str1,str2,str3;
 6 int dfs(int i1,int i2,int p){
 7     if(p==str3.length()){
 8         return 1;
 9     }
10     if( !vis[i1][i2]  ) return 0;
11     
12     vis[i1][i2]=0;
13     
14     if(vis[i1][i2] ==1 ||str1[i1]==str3[p]&&dfs(i1+1,i2,p+1)){
15         vis[i1][i2]=1;
16         return 1;
17     }
18     
19     if(str2[i2]==str3[p]&&dfs(i1,i2+1,p+1)){
20         vis[i1][i2]=1;
21         return 1;
22     }
23     
24     return 0;
25     
26 }
27 int main(){
28     int T,i=1;
29     cin>>T;
30     while(T--){
31         cin>>str1>>str2>>str3;
32         memset(vis,-1,sizeof(vis));
33         
34         if(dfs(0,0,0))
35             cout<<"Data set "<<i<<": yes"<<endl;
36         else
37             cout<<"Data set "<<i<<": no"<<endl;
38         i++;
39     }
40 } 

View Code

主要区别是TLE的代码的标志数组vis[i][j]只能判断在i与j情况下可行；

而ac的代码可以判定在i与j情况下可行或不可行；这样的剪枝才有效（可行的情况只是少部分）。。。

转载于:https://www.cnblogs.com/z-bear/p/8418277.html