两段英文，比较最多有多少个单词连续一样，并将其输出……
由于要求从头打出……dfs一发把……
#include<iostream>
#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
bool flag;
int dp[110][110];
string a[110],b[110];
void out(int i,int j)
{
if(dp[i][j]==0)
return;
if(dp[i][j]==dp[i-1][j-1]+1&&a[i-1]==b[j-1])
{
out(i-1,j-1);
if(flag)
cout<<" ";
cout<<a[i-1];
flag=1;
}
else if(dp[i][j]==dp[i-1][j])
out(i-1,j);
else
out(i,j-1);
}
int main()
{
int i,j,n,m;
string s;
while(cin>>s)
{
for(n=0;s!="#";n++)
{
a[n]=s;
cin>>s;
}
cin>>s;
for(m=0;s!="#";m++)
{
b[m]=s;
cin>>s;
}
memset(dp,0,sizeof(dp));
for(i=1;i<=n;i++)
for(j=1;j<=m;j++)
if(a[i-1]==b[j-1])
dp[i][j]=dp[i-1][j-1]+1;
else
dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
flag=0;
out(n,m);
if(flag)
cout<<endl;
}
return 0;
}
| Time Limit:3000MS | Memory Limit:Unknown | 64bit IO Format:%lld & %llu |
Description
Compromise
| 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
Source
Volume 5. Dynamic Programming
Root :: Competitive Programming: Increasing the Lower Bound of Programming Contests (Steven & Felix Halim) :: Chapter 6. String Processing ::
String Processing with DP
Root :: Competitive Programming 3: The New Lower Bound of Programming Contests (Steven & Felix Halim) :: String Processing :: String Processing with Dynamic Programming ::
Classic
Root :: Competitive Programming 2: This increases the lower bound of Programming Contests. Again (Steven & Felix Halim) :: String Processing ::
String Processing with Dynamic Programming - Standard