HZAU校赛F题 LCS （dp）


17: LCS

Time Limit: 1 Sec  Memory Limit: 128 MB

Submit: 170  Solved: 33

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Description
   Giving two strings consists of only lowercase letters, find the LCS(Longest Common Subsequence) whose all partition are not less than k in length.

Input

There are multiple test cases. In each test case, each of the first two lines is a string(length is less than 2100). The third line is a positive integer k. The input will end by EOF.

Output

    For each test case, output the length of the two strings’ LCS.

Sample Input

abxccdef

abcxcdef

3

abccdef

abcdef

3

Sample Output

4

6

HINT

   In the first test case, the answer is:

      abxccdef

      abcxcdef

    The length is 4.

   In the second test case, the answer is:

      abccdef

      abc def

   The length is 6

Source

题意：求两个字符串里面的LCS，这个LCS的在每个字符串里面的分割长度都要大于等于k

题解：什么叫LCS的分割长度大于等于k呢，就是在两个串里面找长度大于等于k的一样的不重叠的子串，然后长度加起来

普通的LCS是n^2的转移，但是可以不连续，这个的话每段必须要连续至少等于k

我们同样设状态dp[i][j]为到达a[i],b[j]处，这个题目要求的LCS是多少

想想如果a[i]!=b[j],dp[i][j]=max(dp[i-1][j],dp[i][j-1]),是这没有问题的，如果a[i]==b[j]呢，这会就得考虑要不要取这个相等了，如果不取和前面一样，如果要取，那么往前至少得连续k个，开头我想的n^3的方法，就是往前找k个，但是这样不太靠谱，肯定TLE

所以需要预处理出来a[i]b[j]往前有多少个相同的,记为f[i][j]，这个O（n^2）预处理就行了，然后就是如果a[i]==b[j],并且f[i][j]>=k的时候，你可以连续取k个，也可以连续取f[i][j]个，为什么呢，假如你连续k+1个，k>1，然后你取k个，前面连续的不满k个，就直接为0了，所以这时你需要取f[i][j]个，但是可能前面这f[i][j]个不需要连续取，取k个即可能有更有解，所以也能取k个，所以转移为dp[i][j]=max(max(dp[i-1][j],dp[i][j-1]),max(dp[i-k][j-k]+k,dp[i-f[i][j]][j-f[i][j]]+f[i][j]))

#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <sstream>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#pragma comment(linker,"/STACK:102400000,102400000")

using namespace std;
#define MAX 2105
#define MAXN 1000005
#define maxnode 10
#define sigma_size 2
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lrt rt<<1
#define rrt rt<<1|1
#define middle int m=(r+l)>>1
#define LL long long
#define ull unsigned long long
#define mem(x,v) memset(x,v,sizeof(x))
#define lowbit(x) (x&-x)
#define pii pair<int,int>
#define bits(a) __builtin_popcount(a)
#define mk make_pair
#define limit 10000

//const int prime = 999983;
const int INF = 0x3f3f3f3f;
const LL INFF = 0x3f3f;
const double pi = acos(-1.0);
const double inf = 1e18;
const double eps = 1e-9;
const LL mod = 1e9+7;
const ull mxx = 1333331;

/*****************************************************/
inline void RI(int &x) {
char c;
while((c=getchar())<'0' || c>'9');
x=c-'0';
while((c=getchar())>='0' && c<='9') x=(x<<3)+(x<<1)+c-'0';
}
/*****************************************************/

char a[MAX],b[MAX];
int dp[MAX][MAX];
int f[MAX][MAX];

int main(){
//freopen("in.txt","r",stdin);
int k;
while(~scanf("%s%s%d",a+1,b+1,&k)){
int n=strlen(a+1),m=strlen(b+1);
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
if(a[i]==b[j]){
f[i][j]=f[i-1][j-1]+1;
}
else f[i][j]=0;
}
}
mem(dp,0);
for(int i=k;i<=n;i++){
for(int j=k;j<=m;j++){
if(f[i][j]>=k){
dp[i][j]=max(max(dp[i-1][j],dp[i][j-1]),max(dp[i-f[i][j]][j-f[i][j]]+f[i][j],dp[i-k][j-k]+k));
}
else{
dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
}
//cout<<i<<" "<<j<<" "<<dp[i][j]<<endl;
}
}
cout<<dp
[m]<<endl;
}
return 0;
}