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poj 2192 Zipper
题目链接：http://poj.org/problem?id=2192
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 18658 | Accepted: 6651 |
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".
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.
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.
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
题目大意：
给出两串，从两个串取出字符重新组合，看能否组成第三个串。要求：从第一个串取出的字符在第三个串中的顺序不变，第二个串取出的字符在第三个串中的顺序也不变。
算法分析：
此题深搜和DP都能解决：
深搜的话需要几个强有力剪枝条件
1、 第三个串最后一个字符要么是串1的最后一个字符，要么是串2的最后一个字符
2、 按照串1的顺序对串3进行搜索，若不匹配则该字符必是串2的下一个字符。
参考来源：http://www.cnblogs.com/yu-chao/archive/2012/02/26/2369052.html
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 using namespace std; 5 char first[202],second[202],third[402],Left[401]; 6 int sign[402]; 7 bool flag; 8 int check() 9 { 10 int i,count=0; 11 int k=strlen(third); 12 for(i=0;i<k;i++) 13 if(!sign[i]) Left[count++]=third[i]; 14 Left[count]=‘\0‘; 15 if(strcmp(Left,second)==0) return 1; 16 return 0; 17 } 18 int dfs(int f,int s,int t) 19 { 20 if(f>=strlen(first)) 21 { 22 if(check()) flag=true; 23 return 0; 24 } 25 if(flag) return 0; 26 if(first[f]==third[s]) 27 { 28 sign[s]=1; 29 if(s<strlen(third)) dfs(f+1,s+1,t); 30 sign[s]=0; 31 } 32 else 33 { 34 if(third[s]!=second[t]) return 0;//剪枝2 35 } 36 if(!flag && s<strlen(third)) dfs(f,s+1,t+1); 37 return 0; 38 } 39 int main() 40 { 41 int len1,len2,len3,Case,count=0; 42 scanf("%d",&Case); 43 while(Case--) 44 { 45 count++; 46 flag=false; 47 scanf("%s %s %s",first,second,third); 48 memset(sign,0,sizeof(sign)); 49 50 len1=strlen(first); 51 len2=strlen(second); 52 len3=strlen(third); 53 54 if(len1+len2!=len3) 55 { 56 printf("Data set %d: no\n",count); 57 continue; 58 } 59 if(third[len3-1]!=first[len1-1] && third[len3-1]!=second[len2-1])// 剪枝1 60 { 61 printf("Data set %d: no\n",count); 62 continue; 63 } 64 dfs(0,0,0); 65 if(flag) 66 printf("Data set %d: yes\n",count); 67 else 68 printf("Data set %d: no\n",count); 69 } 70 return 0; 71 }
动规算法：（参考来源）
最优子结构分析：如上例，如果A、B可以组成C，那么，C最后一个字母，必定是 A 或 B 的最后一个字母组成。
C去除除最后一位，就变成是否可以求出 A-1和B 或者 A与B-1 与 是否可以构成 C-1。。。
状态转移方程：
用f[i][j] 表示 表示A前 i 为 和B 前j 位是否可以组成 C的前i+j位
dp[i][j]= (dp[i-1][j]&&(a[i]==c[i+j]))||(dp[i][j-1]&&(b[j]==c[i+j]))
1 #include<stdio.h> 2 #include<string.h> 3 4 char a[201],b[201],c[402]; 5 int la,lb,lc; 6 int dp[201][201]; 7 8 int main() 9 { 10 int ncase; 11 scanf("%d",&ncase); 12 for(int n=1; n<=ncase; n++) { 13 14 a[0]=‘p‘; 15 b[0]=‘p‘; 16 c[0]=‘p‘; 17 18 scanf("%s%s%s",a+1,b+1,c+1); 19 20 la=strlen(a); 21 lb=strlen(b); 22 lc=strlen(c); 23 24 la-=1; 25 lb-=1; 26 27 //处理边界 28 for (int i=1; i<=la; i++) 29 if (a[i]==c[i]) dp[i][0]=1; 30 31 for (int i=1; i<=lb; i++) 32 if (b[i]==c[i]) dp[0][i]=1; 33 //DP 34 for (int i=1; i<=la; i++) 35 for (int j=1; j<=lb; j++) 36 dp[i][j]= (dp[i-1][j]&&(a[i]==c[i+j]))||(dp[i][j-1]&&(b[j]==c[i+j])); 37 38 printf("Data set %d: ",n); 39 if (dp[la][lb]==1) printf("yes\n"); 40 else printf("no\n"); 41 42 } 43 }
虽然代码简洁明了易理解，但是个人感觉下面的代码更准确一些。
来源：http://www.cnblogs.com/yu-chao/archive/2012/02/26/2369052.html
若用DP来作先定义res[i][j]=1表示串1前i个字符和串2的前j个字符能组成串3的前i+j个字符，res[i][j]=0则不能。
状态转移方程如下：
Res[i][j]= (third[i+j]==first[i] && res[i-1][j]==1) ||(third[i+j]==second[j]&&res[i][j-1]==1)
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 using namespace std; 5 char first[201],second[201],third[401]; 6 int res[201][201]; 7 int init(int n,int m) 8 { 9 int i; 10 for(i=1;i<=m;i++) 11 if(second[i]==third[i]) res[0][i]=1; 12 else break; 13 for(i=1;i<=n;i++) 14 if(first[i]==third[i]) res[i][0]=1; 15 else break; 16 return 0; 17 } 18 int dp(int n,int m) 19 { 20 int i,j; 21 for(i=1;i<=n;i++) 22 for(j=1;j<=m;j++) 23 { 24 if(third[i+j]==first[i] && res[i-1][j]) res[i][j]=1; 25 if(third[i+j]==second[j] && res[i][j-1]) res[i][j]=1; 26 } 27 if(res[n][m]) return 1; 28 return 0; 29 } 30 int main() 31 { 32 int n,len1,len2,count=0;; 33 scanf("%d",&n); 34 while(n--) 35 { 36 count++; 37 scanf("%s %s %s",first+1,second+1,third+1); 38 len1=strlen(first+1); 39 len2=strlen(second+1); 40 memset(res,0,sizeof(res)); 41 init(len1,len2); 42 43 if(dp(len1,len2)) 44 printf("Data set %d: yes\n",count); 45 else 46 printf("Data set %d: no\n",count); 47 } 48 return 0; 49 }
poj 2192 Zipper
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