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[n][m]<<endl;    }    return 0;}