双指针(最大删除子串)Codeforces Round #579 (Div. 3)--Remove the Substring (hard version)

题目链接:https://codeforces.com/contest/1203/problem/D2

题意:

给你S串、T串,问你最长删除多长的子串使得S串里仍然有T的子序列。

思路:

想了好久,先正着跑一下S串,记录T串每一个字符最左边在哪里,再倒着跑一下,记录T串的每一个字符最右边在哪里。

最后跑一下答案:

1. 开头和结尾特判一下,但不是max( L[1]-1 , l1-R[l2] ) , 而是对两个max( L[1]-1 , l1-L[l2]-1 )、max( R[1]-1 , l1-R[l2]-1 )再取max。

2. 对于中间部分:R[i]-L[i-1]-1 。

  1 #define IOS ios_base::sync_with_stdio(0); cin.tie(0);
  2 #include <cstdio>//sprintf islower isupper
  3 #include <cstdlib>//malloc  exit strcat itoa system("cls")
  4 #include <iostream>//pair
  5 #include <fstream>
  6 #include <bitset>
  7 //#include <map>
  8 //#include<unordered_map>   https://codeforces.com/contest/1203/problem/D2
  9 #include <vector>
 10 #include <stack>
 11 #include <set>
 12 #include <string.h>//strstr substr
 13 #include <string>
 14 #include <time.h>//srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9;
 15 #include <cmath>
 16 #include <deque>
 17 #include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less
 18 #include <vector>//emplace_back
 19 //#include <math.h>
 20 //#include <windows.h>//reverse(a,a+len);// ~ ! ~ ! floor
 21 #include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare)
 22 using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation
 23 #define fo(a,b,c) for(register int a=b;a<=c;++a)
 24 #define fr(a,b,c) for(register int a=b;a>=c;--a)
 25 #define mem(a,b) memset(a,b,sizeof(a))
 26 #define pr printf
 27 #define sc scanf
 28 #define ls rt<<1
 29 #define rs rt<<1|1
 30 void swapp(int &a,int &b);
 31 double fabss(double a);
 32 int maxx(int a,int b);
 33 int minn(int a,int b);
 34 int Del_bit_1(int n);
 35 int lowbit(int n);
 36 int abss(int a);
 37 //const long long INF=(1LL<<60);
 38 const double E=2.718281828;
 39 const double PI=acos(-1.0);
 40 const int inf=(1<<29);
 41 const double ESP=1e-9;
 42 const int mod=(int)1e9+7;
 43 const int N=(int)1e6+10;
 44 
 45 int L[N],R[N];
 46 char s[N],t[N];
 47 
 48 int main()
 49 {
 50 //    freopen("C:\\Users\\13606\\Desktop\\草稿.txt","r",stdin);
 51     int l1,l2;
 52     s[0]=t[0]='$';
 53 //    while(sc("%s%s",s+1,t+1)==2)
 54 //    {
 55         sc("%s%s",s+1,t+1);
 56         l1=strlen(s)-1;
 57         l2=strlen(t)-1;
 58         for(int i=1,pos=1;i<=l1&&pos<=l2;++i)
 59         {
 60             if(t[pos]==s[i])
 61                 L[pos]=i,pos++;
 62         }
 63         for(int i=l1,pos=l2;i>=1&&pos>=1;--i)
 64         {
 65             if(t[pos]==s[i])
 66                 R[pos]=i,pos--;
 67         }
 68         int ans=0;
 69         ans=maxx(ans,maxx(L[1]-1,R[1]-1));
 70         ans=maxx(ans,maxx(l1-L[l2],l1-R[l2]));
 71         for(int i=2;i<=l2;++i)
 72             ans=maxx(ans,R[i]-L[i-1]-1);
 73         pr("%d\n",ans);
 74 //    }
 75     return 0;
 76 }
 77 
 78 /**************************************************************************************/
 79 
 80 int maxx(int a,int b)
 81 {
 82     return a>b?a:b;
 83 }
 84 
 85 void swapp(int &a,int &b)
 86 {
 87     a^=b^=a^=b;
 88 }
 89 
 90 int lowbit(int n)
 91 {
 92     return n&(-n);
 93 }
 94 
 95 int Del_bit_1(int n)
 96 {
 97     return n&(n-1);
 98 }
 99 
100 int abss(int a)
101 {
102     return a>0?a:-a;
103 }
104 
105 double fabss(double a)
106 {
107     return a>0?a:-a;
108 }
109 
110 int minn(int a,int b)
111 {
112     return a<b?a:b;
113 }

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转载自www.cnblogs.com/--HPY-7m/p/11356777.html
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