标签：分析 bsp using NPU cin enter mes 题目 exist
Common Subsequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 55301 Accepted Submission(s): 25537
Problem 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.
The program input is from a text file. Each
data set in the file contains two strings representing the given
sequences. The sequences are separated by any number of white spaces.
The input data are correct. 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
题目大意与分析
就是求最长公共子序列，动态规划即可。
dp[i][j]代表A序列前i-1个元素与B序列前j-1个元素的最长公共子序列的个数，两层循环，相等就++，否则取当前最大的
代码
#include<bits/stdc++.h>
using namespace std;
string x,y;
int i,j,dp[1005][1005];
int main()
{
while(cin>>x>>y)
{
memset(dp,0,sizeof(dp));
for(i=0;i<x.size();i++)
{
for(j=0;j<y.size();j++)
{
if(x[i]==y[j])
dp[i+1][j+1]=dp[i][j]+1;
else
dp[i+1][j+1]=max(dp[i+1][j],dp[i][j+1]);
}
}
cout<<dp[x.size()][y.size()]<<endl;
}
}
HDU 1159 Common Subsequence （动态规划、最长公共子序列）
标签：分析 bsp using NPU cin enter mes 题目 exist
原文地址：https://www.cnblogs.com/dyhaohaoxuexi/p/11392138.html
