LCS(打印全路径) POJ 2264 Advanced Fruits

题目传送门

题意：两个字符串结合起来，公共的字符只输出一次

分析：LCS，记录每个字符的路径

代码：

/*
    LCS(记录路径)模板题：
            用递归打印路径:)
*/
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <iostream>
using namespace std;

const int N = 1e2 + 10;
const int INF = 0x3f3f3f3f;
char s[N], t[N];
int dp[N][N];
int fa[N][N];

void print(int x, int y)　　{
    if (!x && !y)   return ;
    if (fa[x][y] == 0)  {
        print (x-1, y-1);    printf ("%c", s[x-1]);
    }
    else if (fa[x][y] == -1)    {
        print (x-1, y);    printf ("%c", s[x-1]);
    }
    else    {
        print (x, y-1);    printf ("%c", t[y-1]);
    }
}


void LCS(void)  {
    int lens = strlen (s), lent = strlen (t);
    memset (dp, 0, sizeof (dp));
    memset (fa, 0, sizeof (fa));
    for (int i=0; i<=lens; ++i)    fa[i][0] = -1;
    for (int i=0; i<=lent; ++i)    fa[0][i] = 1;

    for (int i=1; i<=lens; ++i) {
        for (int j=1; j<=lent; ++j) {
            if (s[i-1] == t[j-1])   {
                dp[i][j] = dp[i-1][j-1] + 1;
                fa[i][j] = 0;
            }
            else if (dp[i-1][j] >= dp[i][j-1])  {
                dp[i][j] = dp[i-1][j];
                fa[i][j] = -1;
            }
            else    {
                dp[i][j] = dp[i][j-1];
                fa[i][j] = 1;
            }
        }
    }

    // printf ("%d
", dp[lens][lent]);
    print (lens, lent); puts ("");
}


int main(void)  {
    while (scanf ("%s %s", &s, &t) == 2)    {
        LCS ();
    }

    return 0;
}