Light OJ 1110 - An Easy LCS （LCS+字典序最小路径记录）

传送门：http://lightoj.com/volume_showproblem.php?problem=1110
http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=26825

题目大意：求出两个字符串字典序最小的最长公共子序列。

解题思路：过程与求O（n*m）求LCS一样，转移时候加入更新转移当前字典序最小的LCS即可。

Code：

/*   W          w           w        mm          mm             222222222       7777777777777    */
/*    W        w w         w        m  m        m  m          222        22              7777    */
/*    w        w w         w        m  m        m  m                     22              777     */
/*     w      w   w       w        m    m      m    m                    22              77      */
/*     w      w    w      w        m    m      m    m                 222                77      */
/*      w    w      w    w        m      m    m      m              222                  77      */
/*      w    w      w    w        m      m    m      m            222                    77      */
/*       w  w        w  w        m        m  m        m         222                      77      */
/*       w  w        w  w        m        m  m        m      222                         77      */
/*        ww          ww        m          mm          m     222222222222222             77      */

//#pragma comment(linker, "/STACK:102400000,102400000")
//C++
//int size = 256 << 20; // 256MB
//char *p = (char*)malloc(size) + size;
//__asm__("movl %0, %%esp\n" :: "r"(p));
//G++
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<ctime>
#include<deque>
#include<cmath>
#include<vector>
#include<string>
#include<cctype>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<sstream>
#include<iostream>
#include<algorithm>
#define REP(i,s,t) for(int i=(s);i<=(t);i++)
#define REP2(i,t,s) for(int i=(t);i>=(s);i--)

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef unsigned long ul;

int T;
const int N=110;
char a[N],b[N];
int dp[N][N];
char s[N][N][N];
int x,y;
void debug(int x,int y)
{
    printf("\n ");
    REP(i,0,y-1)
    {
        printf(" %c",b[i]);
    }
    printf("\n");
    REP(i,1,x)
    {
        printf("%c",a[i-1]);
        REP(j,1,y)
        {
            printf(" %d",dp[i][j]);
        }
        printf("\n");
    }
    printf("\n");
}
void solve()
{
    x=strlen(a);
    y=strlen(b);
    memset(s,0,sizeof(s));
    memset(dp,0,sizeof(dp));
    REP(i,0,x-1)
    {
        REP(j,0,y-1)
        {
            if(a[i]==b[j])
            {
                dp[i+1][j+1]=dp[i][j]+1;
                REP(k,0,dp[i][j]-1)
                {
                    s[i+1][j+1][k]=s[i][j][k];
                }
                s[i+1][j+1][dp[i][j]]=a[i];
            }
            else
            {
                if(dp[i+1][j]>dp[i][j+1])
                {
                    dp[i+1][j+1]=dp[i+1][j];
                    strcpy(s[i+1][j+1],s[i+1][j]);
                }
                else if(dp[i+1][j]<dp[i][j+1])
                {
                    dp[i+1][j+1]=dp[i][j+1];
                    strcpy(s[i+1][j+1],s[i][j+1]);
                }
                else
                {
                    dp[i+1][j+1]=dp[i][j+1];
                    if(strcmp(s[i+1][j],s[i][j+1])<0)
                    {
                        strcpy(s[i+1][j+1],s[i+1][j]);
                    }
                    else
                    {
                        strcpy(s[i+1][j+1],s[i][j+1]);
                    }
                }
            }
        }
    }
}
int main()
{
  #ifdef ONLINE_JUDGE
  #else
    freopen("test.in","r",stdin);
  #endif
  int ca=1;
  scanf("%d",&T);
  while(T--)
  {
    scanf("%s%s",a,b);
    solve();
    printf("Case %d: ",ca++);
    if(dp[x][y]==0)
    {
        printf(":(\n");
        continue;
    }
    else
    {
        printf("%s\n",s[x][y]);
    }
    //debug(x,y);
  }
  return 0;
}