国产探花免费观看_亚洲丰满少妇自慰呻吟_97日韩有码在线_资源在线日韩欧美_一区二区精品毛片,辰东完美世界有声小说,欢乐颂第一季,yy玄幻小说排行榜完本

首頁 > 學(xué)院 > 開發(fā)設(shè)計(jì) > 正文

Common Subsequence [dp]

2019-11-11 00:28:19
字體:
供稿:網(wǎng)友

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 abfcab programming contest abcd mnp

Sample Output

4 2 0

題解

dp入門題

#include<stdio.h>#include<string.h>#include<algorithm>#define MAX_N 500using namespace std;int dp[2][MAX_N];char a[MAX_N],b[MAX_N];int main(){ while(~scanf("%s%s",a+1,b+1)){ int A=strlen(a+1); int B=strlen(b+1); memset(dp,0,sizeof(dp)); for(int j=1;j<=A;j++) for(int k=1;k<=B;k++){ if(a[j]==b[k]) dp[j&1][k]=dp[(j-1)&1][k-1]+1; else dp[j&1][k]=max(dp[(j-1)&1][k],dp[j&1][k-1]); } printf("%d/n",dp[A&1][B]); } return 0;}
發(fā)表評(píng)論 共有條評(píng)論
用戶名: 密碼:
驗(yàn)證碼: 匿名發(fā)表
主站蜘蛛池模板: 荃湾区| 济阳县| 徐闻县| 巴塘县| 新乡市| 富蕴县| 鹤壁市| 登封市| 稷山县| 津南区| 鹤山市| 龙岩市| 怀宁县| 巨野县| 乐至县| 视频| 花垣县| 旺苍县| 宣恩县| 凤台县| 慈利县| 红河县| 易门县| 柘城县| 上饶市| 揭东县| 成都市| 额尔古纳市| 海城市| 岑溪市| 山东| 同仁县| 镇沅| 岳阳市| 潮安县| 沈阳市| 象州县| 彭水| 永安市| 璧山县| 通许县|