LCS(打印路径) POJ 2250 Compromise

题目传送门

题意：求单词的最长公共子序列，并要求打印路径

分析：LCS 将单词看成一个点，dp[i][j] = dp[i-1][j-1] + 1 (s1[i] == s2[j]), dp[i][j] = max (dp[i-1][j], dp[i][j-1])

代码：

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <string>
using namespace std;

const int N = 1e2 + 10;
const int M = 50;
const int INF = 0x3f3f3f3f;
char s1[M][N], s2[M][N];
int dp[N][N];
int fa[N][N];
bool fir;

void print(int x, int y)    {
    if (!x && !y)    return ;

    if (fa[x][y] == 0)  {
        print (x-1, y-1);
        if (fir)    {
            printf ("%s", s1[x-1]);    fir = false;
        }
        else    printf (" %s", s1[x-1]);
    }
    else if (fa[x][y] == -1)    print (x-1, y);
    else    print (x, y-1);
}

void LCS(int len1, int len2)    {
    memset (dp, 0, sizeof (dp));
    memset (fa, 0, sizeof (fa));
    for (int i=0; i<=len1; ++i)    fa[i][0] = -1;
    for (int i=0; i<=len2; ++i)    fa[0][i] = 1;

    for (int i=1; i<=len1; ++i) {
        for (int j=1; j<=len2; ++j) {
            if (!strcmp (s1[i-1], s2[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;
            }
        }
    }

    fir = true;
    print (len1, len2);        puts ("");
}

int main(void)  {
    while (scanf ("%s", s1[0]) == 1)   {
        int c1 = 1, c2 = 1;
        while (strcmp (s1[c1-1], "#"))    scanf ("%s", &s1[c1++]);
        c1--;   scanf ("%s", s2[0]);
        while (strcmp (s2[c2-1], "#"))    scanf ("%s", &s2[c2++]);
        LCS (c1, c2);
    }

    return 0;
}