HDU 1159(动态规划-最长公共子序列)

问题描述:

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. 

Input

abcfbc abfcabprogramming contest abcd mnp

Output

420

题目题意:求出俩个串的最长公共子序列

题目分析:dp[i][j]表示串s0到si  与t0到tj的最长公共子序列

if (s[i]==t[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]);

代码如下:

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>using namespace std;const int maxn=1e3+100;char s[maxn],t[maxn];int dp[maxn][maxn];int main(){    while (scanf("%s%s",s,t)!=EOF) {        memset (dp,0,sizeof (dp));        int n=strlen(s),m=strlen(t);        for (int i=0;i<n;i++) {            for (int j=0;j<m;j++) {                if (s[i]==t[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]);                }            }        }        printf("%d\n",dp[n][m]);    }    return 0;}