hdu 4681 String（dp）

String

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

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.
  
  

 

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.

  
  

 

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".
   
   
   
   

    
    左右分别跑一遍最长公共子序列，再枚举在A,B中C的位置，然后三段求和即可。复杂度O(n^2)。
   
   
   
   

    
    
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<vector>
#include<cstring>
#include<queue>
#include<stack>
using namespace std;
const int maxn = 1000 + 5;
const int INF = 2000000000;
typedef  pair<int, int> P;
typedef long long LL;

char a[maxn],b[maxn],c[maxn];
int dpl[maxn][maxn],dpr[maxn][maxn];
int n,m;
void pre(){
    n = strlen(a+1);
    m = strlen(b+1);
    memset(dpl,0,sizeof(dpl));
    memset(dpr,0,sizeof(dpr));
    for(int i = 0;i <= m;i++){
        dpl[0][i] = 0;
    }
    for(int i = 0;i <= n;i++){
        dpl[i][0] = 0;
    }
    for(int i = 1;i <= n;i++){
        for(int j = 1;j <= m;j++){
            if(a[i] == b[j]) dpl[i][j] = dpl[i-1][j-1]+1;
            else dpl[i][j] = max(dpl[i-1][j],dpl[i][j-1]);
        }
    }
    for(int i = 0;i <= m;i++){
        dpr[n+1][i] = 0;
    }
    for(int i = 0;i <= n;i++){
        dpr[i][m+1] = 0;
    }
    for(int i = n;i >= 1;i--){
        for(int j = m;j >= 1;j--){
            if(a[i] == b[j]) dpr[i][j] = dpr[i+1][j+1]+1;
            else dpr[i][j] = max(dpr[i+1][j],dpr[i][j+1]);
        }
    }
}

int la[maxn],lb[maxn],ra[maxn],rb[maxn];

int main(){
    int t,kase = 0;
    scanf("%d",&t);
    while(t--){
        kase++;
        scanf("%s",a+1);
        scanf("%s",b+1);
        scanf("%s",c+1);
        pre();
        int len = strlen(c+1);
        int p = 1,cnta = 0;
        while(p <= n){
            int fir = -1;
            int i,j;
            for(i = 1,j = p;i <= len && j <= n;){
                if(c[i] == a[j]){
                    if(fir == -1) fir = j;
                    i++;
                }
                j++;
            }
            p++;
            if(i == len+1){
               la[cnta] = fir;
               ra[cnta++] = j-1;
            }
        }
        p = 1;
        int cntb = 0;
        while(p <= m){
            int fir = -1;
            int i,j;
            for(i = 1,j = p;i <= len && j <= m;){
                if(c[i] == b[j]){
                    if(fir == -1) fir = j;
                    i++;
                }
                j++;
            }
            p++;
            if(i == len+1){
               lb[cntb] = fir;
               rb[cntb++] = j-1;
            }
        }
        int ans = 0;
        for(int i = 0;i < cnta;i++){
            for(int j = 0;j < cntb;j++){
                ans = max(ans,len+dpl[la[i]-1][lb[j]-1]+dpr[ra[i]+1][rb[j]+1]);
            }
        }
        printf("Case #%d: %d\n",kase,ans);
    }
    return 0;
}