HDU 4679 String
String
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 695 Accepted Submission(s): 254
Total Submission(s): 695 Accepted Submission(s): 254
Problem Description
Given 3 strings A, B, C, find the longest string D which satisfy the following rules:
a) D is the subsequence of A
b) D is the subsequence of B
c) C is the substring of D
Substring here means a consecutive subsequnce.
You need to output the length of D.
a) D is the subsequence of A
b) D is the subsequence of B
c) C is the substring of D
Substring here means a consecutive subsequnce.
You need to output the length of D.
Input
The first line of the input contains an integer T(T = 20) which means the number of test cases.
For each test case, the first line only contains string A, the second line only contains string B, and the third only contains string C.
The length of each string will not exceed 1000, and string C should always be the subsequence of string A and string B.
All the letters in each string are in lowercase.
For each test case, the first line only contains string A, the second line only contains string B, and the third only contains string C.
The length of each string will not exceed 1000, and string C should always be the subsequence of string A and string B.
All the letters in each string are in lowercase.
Output
For each test case, output Case #a: b. Here a means the number of case, and b means the length of D.
Sample Input
2 aaaaa aaaa aa abcdef acebdf cf
Sample Output
Case #1: 4 Case #2: 3
Hint
For test one, D is "aaaa", and for test two, D is "acf".
Source
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zhuyuanchen520
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#include <iostream>
#include <string>
#include <cstring>
#include <cstdio>
#define N 1100
using namespace std;
string s1,s2,s3,s4,s5;
char temp[N];
int num1[N][N],num2[N][N];
struct num
{
int sta,end;
} a[N],b[N];
int main()
{
//freopen("data.in","r",stdin);
void get(int (*p)[N],string ch1,string ch2);
int t,tem=1;
scanf("%d",&t);
while(t--)
{
cin>>s1>>s2>>s3;
get(num1,s1,s2);
int l1 = s1.size();
for(int i=0; i<=l1-1; i++)
{
temp[l1-1-i] = s1[i];
}
temp[l1] = '\0';
s4 = temp;
int l2 = s2.size();
for(int i=0; i<=l2-1; i++)
{
temp[l2-1-i] = s2[i];
}
temp[l2] = '\0';
s5 = temp;
get(num2,s4,s5);
int l3 = s3.size(),Top1=0,Top2=0;
for(int i=0; i<=l1-1; i++)
{
if(s1[i]==s3[0])
{
int x = 0;
int sta = i;
int end = -1;
for(int j=i; j<=l1-1&&x<=l3-1; j++)
{
if(s1[j]==s3[x])
{
x++;
}
if(x==l3)
{
end = j;
}
}
if(end==-1)
{
continue;
}
a[Top1].sta = sta;
a[Top1++].end = end;
}
}
for(int i=0; i<=l2-1; i++)
{
if(s2[i]==s3[0])
{
int x = 0;
int sta = i;
int end = -1;
for(int j=i; j<=l2-1&&x<=l3-1; j++)
{
if(s2[j]==s3[x])
{
x++;
}
if(x==l3)
{
end = j;
}
}
if(end==-1)
{
continue;
}
b[Top2].sta = sta;
b[Top2++].end = end;
}
}
int res = l3,Max=0;
for(int i=0; i<=Top1-1; i++)
{
for(int j=0; j<=Top2-1; j++)
{
int x1 = a[i].sta;
int y1 = a[i].end;
int x2 = b[j].sta;
int y2 = b[j].end;
int k1 = 0;
if(x1>0&&x2>0)
{
k1 = num1[x1-1][x2-1];
}
int k2 = 0;
if(y1<l1-1&&y2<l2-1)
{
k2 = num2[l1-2-y1][l2-2-y2];
}
Max = max(Max,res+k2+k1);
}
}
printf("Case #%d: %d\n",tem++,Max);
}
return 0;
}
void get(int (*p)[N],string ch1,string ch2)
{
int l1 = ch1.size();
int l2 = ch2.size();
memset(p,0,sizeof(p));
for(int i=0; i<=l1-1; i++)
{
for(int j=0; j<=l2-1; j++)
{
if(i==0&&j==0)
{
if(ch1[i]==ch2[j])
{
p[i][j] = 1;
}
}
else if(i==0&&j!=0)
{
if(ch1[i]==ch2[j])
{
p[i][j] = 1;
}
else
{
p[i][j] = p[i][j-1];
}
}
else if(i!=0&&j==0)
{
if(ch1[i]==ch2[j])
{
p[i][j] = 1;
}
else
{
p[i][j] = p[i-1][j];
}
}
else
{
if(ch1[i]==ch2[j])
{
p[i][j] = p[i-1][j-1]+1;
}
else
{
p[i][j] = max(p[i][j-1],p[i-1][j]);
}
}
}
}
}