标签：source test search pst scan std other char output
| Time Limit: 1000MS |
|
Memory Limit: 10000K |
| Total Submissions: 53247 |
|
Accepted: 22084 |
Description
A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = < x1, x2, ..., xm > another sequence Z = < z1, z2, ..., zk > is a subsequence of X if there exists a strictly increasing sequence < i1, i2, ..., ik > of indices of X such that for all j = 1,2,...,k, xij = zj. For example, Z = < a, b, f, c > is a subsequence of X = < a, b, c, f, b, c > with index sequence < 1, 2, 4, 6 >. Given two sequences X and Y the problem is to find the length of the maximum-length common subsequence of X and Y.
Input
The program input is from the std input. Each data set in the input contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct.
Output
For each set of data the program prints on the standard output the length of the maximum-length common subsequence from the beginning of a separate line.
Sample Input
abcfbc abfcab
programming contest
abcd mnp
Sample Output
4
2
0
Source
用dp[n][m]表示第一个喘选到n，第二个串选到m
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<cmath>
5 #include<algorithm>
6 using namespace std;
7 const int MAXN=601;
8 const int maxn=0x7fffff;
9 char a[MAXN];
10 char b[MAXN];
11 int dp[MAXN][MAXN];
12 int la,lb;
13 int main()
14 {
15 while(~scanf("%s%s",a,b))
16 {
17 la=strlen(a);
18 lb=strlen(b);
19 memset(dp,0,sizeof(dp));
20 //dp[0][0]=1;
21 for(int i=1;i<=la;i++)
22 {
23 for(int j=1;j<=lb;j++)
24 {
25
26 if(a[i-1]==b[j-1])
27 dp[i][j]=max(dp[i][j],dp[i-1][j-1]+1);
28 else dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
29 }
30 }
31 printf("%d\n",dp[la][lb]);
32 }
33 return 0;
34 }
Common Subsequence
标签：source test search pst scan std other char output
原文地址：http://www.cnblogs.com/zwfymqz/p/7202106.html
