17: LCS
Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 170 Solved: 33
[Submit][Status][Web Board]
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; }