hdu 1159 整理下水题 递推动规
来源:互联网 发布:绝地求生有数据接口吗 编辑:程序博客网 时间:2024/04/29 22:14
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.
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 abfcabprogramming contest abcd mnp
Sample Output
420
#include<stdio.h>#include<string.h>using namespace std;char a[1000],b[1000];int dp[1000][1000];int main(){ while( scanf("%s%s",a,b)!=EOF){ memset(dp,0,sizeof(dp)); int i,j;int l=strlen(a),s=strlen(b); for(i=1;i<=l;i++) for(j=1;j<=s;j++) { if(a[i-1]==b[j-1]) dp[i][j]=dp[i-1][j-1]+1; else dp[i][j]=dp[i-1][j]>dp[i][j-1]?dp[i-1][j]:dp[i][j-1]; } printf("%d\n",dp[l][s]); }}
0 0
- hdu 1159 整理下水题 递推动规
- hdu 2013 整理下水题 递推动规
- hdu 2041 整理下水题 递推动规
- hdu 2569 整理下水题 递推动规
- hdu 2084 整理下水题 递推动规
- hdu 2037 整理下水题
- hdu 1870 整理下水题
- hdu 1711 整理下水题
- hdu 1009 整理下水题 贪心
- hdu 1789 整理下水题 贪心
- hdu 1106 整理下水题 排序
- hdu 2020 整理下水题 排序
- hdu 1425 整理下水题 排序
- hdu 1785 整理下水题 排序
- hdu 2602 整理下水题 背包
- hdu 1702 整理下水题 栈和队列
- hdu 1022 整理下水题 栈和队列
- HDU-1076-An Easy Task(Debian下水题测试.....)
- PHP中WebService应用_SOAP
- c++11 对容器进行排序
- 数据结构算法代码实现——栈和队列(二)
- 剑指offer--面试题16:翻转链表--Java实现
- [java]类加载机制
- hdu 1159 整理下水题 递推动规
- Git详解之三:Git分支
- 小数化分数2
- Bmob
- struts2 java.lang.UnsupportedClassVersionError: Bad version number in .class file 解决
- Android的Menu
- hdu 2602 整理下水题 背包
- 架构师速成2-概述
- 城市修路 连通