「CF477C」Dreamoon and Strings
Dreamoon has a string s and a pattern string p. He first removes exactly x characters from sobtaining string s‘ as a result. Then he calculates 
that is defined as the maximal number of non-overlapping substrings equal to p that can be found in s‘. He wants to make this number as big as possible.
More formally, let’s define 
as maximum value of over all s‘ that can be obtained by removing exactly x characters from s. Dreamoon wants to know for allx from 0 to |s| where |s| denotes the length of string s.
The first line of the input contains the string s (1 ≤ |s| ≤ 2 000).
The second line of the input contains the string p (1 ≤ |p| ≤ 500).
Both strings will only consist of lower case English letters.
Print |s| + 1 space-separated integers in a single line representing the for all x from0 to |s|.
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1 2 |
aaaaa aa |
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1 |
2 2 1 1 0 0 |
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1 2 |
axbaxxb ab |
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1 |
0 1 1 2 1 1 0 0 |
For the first sample, the corresponding optimal values of s‘ after removal 0 through |s| = 5characters from s are {“aaaaa”, “aaaa“, “aaa”, “aa“, “a”, “”}.
For the second sample, possible corresponding optimal values of s‘ are {“axbaxxb”,“abaxxb”, “axbab“, “abab“, “aba”, “ab“, “a”, “”}.
题解
f[i][j]表示前i个删j个可以匹配的最大次数
f[i][j]=max(f[i-1][j],f[i-len(p)-cal()][j-cal()]+1)
cal()表示从第i位往前，使得s的末尾与p匹配最少要删多少个
因为f[i][j]>=f[i-1][j-1]所以多删没有意义
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#include<iostream> #include<cstdio> #include<cstdlib> #include<algorithm> #include<cmath> #include<cstring> #define inf 1000000000 #define ll long long #define mod 1000000007 using namespace std; int read() { int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } int n,len; char s[2005],ch[2005]; int f[2005][2005]; int cal(int x) { if(x<len)return inf; int ans=0,p=len,q=x; while(p&&q) { if(s[q]==ch[p])p--,q--; else ans++,q--; } if(p==0)return ans; return inf; } int main() { scanf("%s%s",s+1,ch+1); n=strlen(s+1),len=strlen(ch+1); for(int i=0;i<=n;i++) for(int j=0;j<=n;j++) if(i<j)f[i][j]=-inf; for(int i=1;i<=n;i++) { int x=cal(i); for(int j=0;j<=n;j++) { f[i][j]=max(f[i][j],f[i-1][j]); if(j-x>=0)f[i][j]=max(f[i][j],f[i-len-x][j-x]+1); } } for(int i=0;i<=n;i++) printf("%d ",f[n][i]); return 0; } |
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