light oj 1159 - Batman LCS

学过简单动态规划的人应该对最长公共子序列的问题很熟悉了，这道题只不过多加了一条字符串变成三条了，还记得，只要把状态变成三维的即可。

//http://lightoj.com/volume_showproblem.php?problem=1159
//2013-08-15-09.50
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string.h>

using namespace std;

char str1[55];
char str2[55];
char str3[55];
int dp[55][55][55];

int maxval(int a, int b, int c)
{
    a = max(a, b);
    a = max(a, c);
    return a;
}
int main()
{
    int t, kase = 0;
    scanf("%d", &t);
    while (t--)
    {
        int ans = 0;
        scanf("%s %s %s", &str1[1], &str2[1], &str3[1]);
        int l1 = strlen(&str1[1]);
        int l2 = strlen(&str2[1]);
        int l3 = strlen(&str3[1]);
        memset(dp, 0, sizeof(dp));
        for (int i = 1; i <= l1; i++)
        {
            for (int j = 1; j <= l2; j++)
            {
                for (int k = 1; k <= l3; k++)
                {
                    if (str1[i] == str2[j] && str2[j] == str3[k])
                        dp[i][j][k] = dp[i-1][j-1][k-1] + 1;
                    else
                    {
                        dp[i][j][k] = maxval(dp[i-1][j][k], dp[i][j-1][k], dp[i][j][k-1]);
                    }
                }
            }
        }
        printf("Case %d: %d\n", ++kase, dp[l1][l2][l3]);
    }
    return 0;
}