poj 2250 Compromise(区间dp)

题目链接：http://poj.org/problem?id=2250

思路分析：最长公共子序列问题的变形，只是把字符变成了字符串，按照最长公共子序列的思路即可以求解。

代码如下： 

#include <stdio.h>
#include <string.h>

#define IsEqual(a, b) strcmp((a), (b)) == 0
enum { Left, Up, UpAndLeft };
int XLen, YLen;
const int MAX_N = 100 + 10;
char X[MAX_N][35], Y[MAX_N][35];
int dp[MAX_N][MAX_N], r[MAX_N][MAX_N];

void PrintWords(int i, int j)
{
    if (i == 0 || j == 0)
        return;

    if (r[i][j] == UpAndLeft)
    {
        PrintWords(i - 1, j - 1);

        if (i == XLen && j == YLen)
            printf("%s",X[i]);
        else
            printf("%s ", X[i]);
    }
    else
    if (r[i][j] == Up)
        PrintWords(i - 1, j);
    else
        PrintWords(i, j - 1);
}

void Lcs( int XLen, int YLen )
{
    for (int i = 1; i <= XLen; ++i)
        dp[i][0] = 0;
    for (int j = 0; j <= YLen; ++j)
        dp[0][j] = 0;

    for (int i = 1; i <= XLen; ++i)
        for (int j = 1; j <= YLen; ++j)
        {
            if (IsEqual(X[i], Y[j]))
            {
                dp[i][j] = dp[i - 1][j - 1] + 1;
                r[i][j] = UpAndLeft;
            }
            else
            if (dp[i - 1][j] >= dp[i][j - 1])
            {
                dp[i][j] = dp[i - 1][j];
                r[i][j] = Up;
            }
            else
            {
                dp[i][j] = dp[i][j - 1];
                r[i][j] = Left;
            }
        }
}

int main()
{
    XLen = YLen = 1;

    while (scanf("%s", X[XLen]) != EOF)
    {
        memset(dp, 0, sizeof(dp));
        memset(r, -1, sizeof(r));

        while (1)
        {
            scanf("%s", X[++XLen]);
            if (strcmp("#", X[XLen]) == 0)
            {
                XLen--;
                break;
            }
        }

        while (1)
        {
            scanf("%s", Y[YLen++]);
            if (strcmp("#", Y[YLen - 1]) == 0)
            {
                YLen -= 2;
                break;
            }
        }

        Lcs(XLen, YLen);
        PrintWords(XLen, YLen);
        printf("
");
        XLen = YLen = 1;
    }

    return 0;
}