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  • HDU 1501 Zipper 动态规划经典

    Zipper

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 4884    Accepted Submission(s): 1742

    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
     

    其实还是LCS算法的思想。

    opt[i][j] 表示 字符串2的前i个字符 和 字符串1 的前j个字符,可以匹配 字符串3 的最大个数。

    例如 对于第一个case(省略第0行和第0列), 参看代码:

    Data set 1: 

    2 3 3 3
    2 4 5 5
    2 4 6 7


    状态方程:

    opt[i][j] = max(opt[i-1][j] + (str1[ i-1 ] == str3[ opt[i-1][j] ]), opt[i][j-1] + (str2[ j-1 ] == str3[ opt[i][j-1] ]) );


    #include <stdio.h>
    #include <string.h>
    #include <iostream>
    using namespace std;
    
    int max(int a,int b){
    	return a > b ? a:b;
    }
    
    int main(){
    	//freopen("in.txt","r",stdin);
    	int t;
    	scanf("%d",&t);
    	char str1[210],str2[210],str3[410];
    	int opt[410][410];
    	for(int c=1; c<=t; c++){
    		printf("Data set %d: ",c);
    		scanf("%s %s %s", str1,str2,str3);
    		int len1 = strlen(str1);
    		int len2 = strlen(str2);
    		int k = 0;
    			opt[0][0] = 0;
    	
    		for(int i=1; i<=len2; i++){
    			if(str2[i-1] == str3[i-1])
    				opt[0][i] = 	opt[0][i-1] + 1;
    			else
    				opt[0][i] = opt[0][i-1];
    		}
    		for( i=1; i<=len1; i++){
    			if(str1[i-1] == str3[i-1])
    				opt[i][0] = opt[i-1][0] + 1;
    			else
    				opt[i][0] =  opt[i-1][0];
    		}
    		
    		for( i=1; i<=len1; i++){
    			for(int j=1; j<=len2; j++){
    				
    				opt[i][j] = max(opt[i-1][j] + (str1[ i-1 ] == str3[ opt[i-1][j] ]), opt[i][j-1] + (str2[ j-1 ] == str3[ opt[i][j-1] ]) );
    				//cout << opt[i][j] << " ";
    			}
    		}
    		if(opt[len1][len2] == len1 + len2){
    			printf("yes
    ");
    		}else
    			printf("no
    ");
    	
    	}
    	return 0;
    }



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  • 原文地址:https://www.cnblogs.com/snake-hand/p/3157393.html
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