light oj 1159 - Batman LCS

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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;}