[LCS]Uva531 Compromise

LCS详解
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Compromise

In a few months the European Currency Union will become a reality. However, to join the club, the Maastricht criteria must be fulfilled, and this is not a trivial task for the countries (maybe except for Luxembourg). To enforce that Germany will fulfill the criteria, our government has so many wonderful options (raise taxes, sell stocks, revalue the gold reserves,…) that it is really hard to choose what to do.
Therefore the German government requires a program for the following task:
Two politicians each enter their proposal of what to do. The computer then outputs the longest common subsequence of words that occurs in both proposals. As you can see, this is a totally fair compromise (after all, a common sequence of words is something what both people have in mind).
Your country needs this program, so your job is to write it for us.
Input Specification
The input file will contain several test cases.
Each test case consists of two texts. Each text is given as a sequence of lower-case words, separated by whitespace, but with no punctuation. Words will be less than 30 characters long. Both texts will contain less than 100 words and will be terminated by a line containing a single ‘#’.
Input is terminated by end of file.
Output Specification
For each test case, print the longest common subsequence of words occuring in the two texts. If there is more than one such sequence, any one is acceptable. Separate the words by one blank. After the last word, output a newline character.
Sample Input
die einkommen der landwirte
sind fuer die abgeordneten ein buch mit sieben siegeln
um dem abzuhelfen
muessen dringend alle subventionsgesetze verbessert werden
#
die steuern auf vermoegen und einkommen
sollten nach meinung der abgeordneten
nachdruecklich erhoben werden
dazu muessen die kontrollbefugnisse der finanzbehoerden
dringend verbessert werden
#
Sample Output
die einkommen der abgeordneten muessen dringend verbessert werden
题意：给定两个文本，要求输出最多相同的单词序列。

思路：LCS，把单词一个一个记录，，strcmp(s1,s2)相同为0，不同为1。。
这道题主要是打印路径，，递归打印，具体看代码

#include<iostream>
#include<cstdio>
#include<ctime>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<string>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<algorithm>
#ifdef WIN32
#define AUTO "%I64d"
#else
#define AUTO "%lld"
#endif
#define INF 0x3f3f3f3f
#define clock CLOCKS_PER_SEC
#define cle(x) memset(x,0,sizeof(x))
#define maxcle(x) memset(x,127,sizeof(x))
#define mincle(x) memset(x,-1,sizeof(x))
#define cop(a,x) memcpy(x,a,sizeof(a))
#define FROP "Uva"
#define C(a,b) next_permutation(a,b)
#define LL long long
#define smin(x,tmp) x=min(x,tmp)
using namespace std;
const int N=110,M=35;
char s1[N][M],s2[N][M];
int f[N][N],n,m,path[N][N];
int ans;
void LCS()
{
    for(int i = 1; i <= n; i++)
        for(int j = 1; j<= m; j++)
        {
            if(!strcmp(s1[i],s2[j]))
                f[i][j]=f[i-1][j-1]+1,path[i][j]=1;
            else if(f[i-1][j]>f[i][j-1])f[i][j]=f[i-1][j],path[i][j]=2;
            else f[i][j]=f[i][j-1],path[i][j]=3;
        }
    ans=f[n][m];
}
void print(int i,int j)
{
    if(!i||!j)return;
    if(path[i][j]==1)
    {
        print(i-1,j-1);
        if(--ans==0)printf("%s\n",s1[i]);
        else printf("%s ",s1[i]);
    }
    else if(path[i][j]==2)print(i-1,j);
    else print(i,j-1);
}
int main()
{
    freopen(FROP".in","r",stdin);
    freopen(FROP".out","w",stdout);
    while(scanf("%s",s1[++n])!=EOF){
        while(scanf("%s",s1[n+1])&&s1[n+1][0]!='#')n++;
        while(scanf("%s",s2[m+1])&&s2[m+1][0]!='#')m++;
        LCS();
        print(n,m);
        n=0,m=0;
    }
    return 0;
}