题目链接:UVa 10192 - Vacation
LCS。
#include <iostream>
#include <stdio.h>
#include <cstring>
using namespace std;
const int MAX_N = 100 + 10;
int dp[MAX_N][MAX_N];
char a[MAX_N],b[MAX_N];
int main()
{
int num = 0;
while(gets(a) && a[0] != '#')
{
gets(b);
int len1 = strlen(a);
int len2 = strlen(b);
for(int i = 0;i <= len1;i++)
dp[i][0] = 0;
for(int i = 1;i <= len2;i++)
dp[0][i] = 0;
for(int i = 1;i <= len1;i++)
{
for(int j = 1;j <= len2;j++)
{
if(a[i - 1] == b[j - 1])
dp[i][j] = dp[i - 1][j - 1] + 1;
else
dp[i][j] = max(dp[i - 1][j],dp[i][j - 1]);
}
}
//if(dp[len1][len2] > 1)
cout << "Case #" << ++num << ": you can visit at most " << dp[len1][len2] << " cities." << endl;
/*else
cout << "Case #" << ++num << ": you can visit at most " << dp[len1][len2] << " city." << endl;*/
}
return 0;
}
本文解决UVa10192-Vacation问题,使用LCS(最长公共子序列)算法,通过输入两段字符串并输出最长公共子序列的长度。程序采用动态规划方法实现,包含字符串读取、动态表初始化和计算过程。
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