CF750E 线段树+矩阵乘矩阵加

题目描述

A string tt is called nice if a string "2017" occurs in tt as a subsequence but a string "2016" doesn't occur in tt as a subsequence. For example, strings "203434107" and "9220617" are nice, while strings "20016", "1234" and "20167" aren't nice.

The ugliness of a string is the minimum possible number of characters to remove, in order to obtain a nice string. If it's impossible to make a string nice by removing characters, its ugliness is -1−1 .

Limak has a string ss of length nn , with characters indexed 11 through nn . He asks you qq queries. In the ii-th query you should compute and print the ugliness of a substring (continuous subsequence) of ssstarting at the index a_{i}ai and ending at the index b_{i}bi (inclusive).

输入输出格式

输入格式：

The first line of the input contains two integers nn and qq ( 4<=n<=2000004<=n<=200000 , 1<=q<=2000001<=q<=200000 ) — the length of the string ss and the number of queries respectively.

The second line contains a string ss of length nn . Every character is one of digits '0'–'9'.

The ii -th of next qq lines contains two integers a_{i}ai and b_{i}bi ( 1<=a_{i}<=b_{i}<=n1<=ai<=bi<=n ), describing a substring in the ii -th query.

输出格式：

For each query print the ugliness of the given substring.

输入输出样例

输入样例#1：

输出样例#1：

输入样例#2：

15 5

012016662091670

3 4

1 14

4 15

1 13

10 15

输出样例#2：

输入样例#3：

输出样例#3：

-1

-1

说明

In the first sample:

In the first query, ugliness(ugliness( "20166766" )=4)=4 because all four sixes must be removed.
In the second query, ugliness(ugliness( "2016676" )=3)=3 because all three sixes must be removed.
In the third query, ugliness(ugliness( "0166766" )=-1)=−1 because it's impossible to remove some digits to get a nice string.

In the second sample:

In the second query, ugliness(ugliness( "01201666209167" )=2)=2 . It's optimal to remove the first digit '2' and the last digit '6', what gives a string "010166620917", which is nice.
In the third query, ugliness(ugliness( "016662091670" )=1)=1 . It's optimal to remove the last digit '6', what gives a nice string "01666209170".

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这是一个大坑

先把代码丢在这里，改天详细写个题解 233333

 #include<iostream>

 #include<cstdio>

 #include<cmath>

 #include<cstring>

 #include<algorithm>

 #define N 200100

 #define INF 1e9

 #define lc (p<<1)

 #define rc (p<<1|1)

 using namespace std;

 int n,m;

 char s[N];

 struct node

 {

     int a[][];

     node(){

         for(int i=;i<;i++)

             for(int j=;j<;j++)

                 a[i][j]=INF;

     }

 }T[N<<];

 char ch[]={'','','','',''};

 int find (char x)//返回数字本身的愚蠢办法

 {

     for(int i=;i<;i++)

         if(x==ch[i]) return i;

     return -;

 }

 node pushup(node &a,node &b){

     node res;

     for(int i=;i<;i++)

         for(int j=i;j<;j++)

             for(int k=i;k<=j;k++)

                 res.a[i][j]=min(res.a[i][j],a.a[i][k]+b.a[k][j]);

     return res;

 }

 void build(int p,int l,int r)

 {

     if(l==r)

     {

         int f=find(s[l]);

         for(int i=;i<;i++)

             T[p].a[i][i]=;//初始化——隔壁有更好的方法

         //以下为转移方程初始化

         if(f!=-&&f<)//l的值为2 0 1 7

         {

             T[p].a[f][f+]=;

             T[p].a[f][f]=;

         }

         else if(f==)//l为6

             T[p].a[][]=T[p].a[][]=;

         return;

     }

     int mid=(l+r)>>;

     build(lc,l,mid);

     build(rc,mid+,r);

     T[p]=pushup(T[lc],T[rc]);

 }

 node query(int p,int l,int r,int ql,int qr)

 {

     if(ql==l&&qr==r)

         return T[p];

     int mid=(l+r)>>;

     if(qr<=mid) return query(lc,l,mid,ql,qr);

     if(ql>mid) return query(rc,mid+,r,ql,qr);

     else

     {

         node tmpl=query(lc,l,mid,ql,mid);

         node tmpr=query(rc,mid+,r,mid+,qr);

         return pushup(tmpl,tmpr);

     }

 }

 int main()

 {

     scanf("%d%d",&n,&m);

     scanf("%s",s+);

     build(,,n);

     while(m--)

     {

         int l,r;

         scanf("%d%d",&l,&r);

         int ans=query(,,n,l,r).a[][];

         if(ans==INF) printf("-1\n");

         else printf("%d\n",ans);

     }

     return ;

 }