标签：and while oss cep 题目 exist %s AMM each
题目链接:http://poj.org/problem?id=1458
Common Subsequence
| Time Limit: 1000MS |
|
Memory Limit: 10000K |
| Total Submissions: 63588 |
|
Accepted: 26535 |
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
裸的LCS,直接做就好
注意多组输入输出
%s以空格就可以分隔，不用非得回车
直接贴代码了
1 #include<cstdio>
2 #include<cstring>
3 #include<string>
4 #include<iostream>
5 #define mem(a,b) memset(a,0,sizeof(a))
6 using namespace std;
7 int dp[10000][10000];
8 char a[10001],b[10001],c[20002];
9 int main()
10 {
11 while(~scanf("%s%s",a,b))
12 {
13 int lena=strlen(a),lenb=strlen(b);
14 for(int i=1;i<=lena;i++)
15 {
16 for(int j=1;j<=lenb;j++)
17 {
18 if(a[i-1]==b[j-1]) dp[i][j]=dp[i-1][j-1]+1;
19 else dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
20 }
21 }
22 printf("%d\n",dp[lena][lenb]);
23 mem(a,0),mem(b,0);
24 }
25 return 0;
26 }
poj1458(裸LCS)
标签：and while oss cep 题目 exist %s AMM each
原文地址：https://www.cnblogs.com/codeoosacm/p/9979297.html
