POJ 1458-Common Subsequence（线性dp/LCS）

Common Subsequence

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 39009 Accepted: 15713

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, x_ij = 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         abfcabprogramming    contest abcd           mnp

Sample Output

420

最长公共子序列问题。

二维dp。dp[i][j]代表字符串s的前i个字符与字符串t的前j个字符的最长公共子序列的长度。dp[0][0]=0;

if(s[i]==t[j])dp[i][j]=dp[i-1][j-1]+1;

else dp[i][j]=max(dp[i][j-1],dp[i-1][j]);//从前状态取最大

数组建议用short。。

#include <algorithm>#include <iostream>#include <cstring>#include <cstdlib>#include <string>#include <cctype>#include <vector>#include <cstdio>#include <cmath>#include <queue>#include <stack>#include <map>#include <set>#define ll long long#define maxn 360#define pp pair<int,int>#define INF 0x3f3f3f3f#define max(x,y) ( ((x) > (y)) ? (x) : (y) )#define min(x,y) ( ((x) > (y)) ? (y) : (x) )using namespace std;short dp[maxn][maxn];char s[maxn], t[maxn];void solve(){memset(dp, 0, sizeof(dp));int ls = strlen(s), lt = strlen(t);for (int i = 1; i <= ls; i++)for (int j = 1; j <= lt; j++) {dp[i][j] = s[i - 1] == t[j - 1] ? dp[i - 1][j - 1] + 1 : max(dp[i - 1][j], dp[i][j - 1]);}printf("%d\n", dp[ls][lt]);}int main(){while (scanf("%s %s", s, t) != EOF) {solve();}return 0;}