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
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
1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #include<cstring> 5 #include<algorithm> 6 #define N 200100 7 #define INF 1e9 8 #define lc (p<<1) 9 #define rc (p<<1|1) 10 using namespace std; 11 int n,m; 12 char s[N]; 13 struct node 14 { 15 int a[5][5]; 16 node(){ 17 for(int i=0;i<5;i++) 18 for(int j=0;j<5;j++) 19 a[i][j]=INF; 20 } 21 }T[N<<2]; 22 char ch[]={'2','0','1','7','6'}; 23 int find (char x)//返回数字本身的愚蠢办法 24 { 25 for(int i=0;i<5;i++) 26 if(x==ch[i]) return i; 27 return -1; 28 } 29 node pushup(node &a,node &b){ 30 node res; 31 for(int i=0;i<5;i++) 32 for(int j=i;j<5;j++) 33 for(int k=i;k<=j;k++) 34 res.a[i][j]=min(res.a[i][j],a.a[i][k]+b.a[k][j]); 35 return res; 36 } 37 void build(int p,int l,int r) 38 { 39 if(l==r) 40 { 41 int f=find(s[l]); 42 for(int i=0;i<5;i++) 43 T[p].a[i][i]=0;//初始化——隔壁有更好的方法 44 45 //以下为转移方程初始化 46 if(f!=-1&&f<4)//l的值为2 0 1 7 47 { 48 T[p].a[f][f+1]=0; 49 T[p].a[f][f]=1; 50 } 51 else if(f==4)//l为6 52 T[p].a[3][3]=T[p].a[4][4]=1; 53 return; 54 } 55 int mid=(l+r)>>1; 56 build(lc,l,mid); 57 build(rc,mid+1,r); 58 T[p]=pushup(T[lc],T[rc]); 59 } 60 node query(int p,int l,int r,int ql,int qr) 61 { 62 if(ql==l&&qr==r) 63 return T[p]; 64 int mid=(l+r)>>1; 65 if(qr<=mid) return query(lc,l,mid,ql,qr); 66 if(ql>mid) return query(rc,mid+1,r,ql,qr); 67 else 68 { 69 node tmpl=query(lc,l,mid,ql,mid); 70 node tmpr=query(rc,mid+1,r,mid+1,qr); 71 return pushup(tmpl,tmpr); 72 } 73 } 74 int main() 75 { 76 scanf("%d%d",&n,&m); 77 scanf("%s",s+1); 78 build(1,1,n); 79 while(m--) 80 { 81 int l,r; 82 scanf("%d%d",&l,&r); 83 int ans=query(1,1,n,l,r).a[0][4]; 84 if(ans==INF) printf("-1\n"); 85 else printf("%d\n",ans); 86 } 87 return 0; 88 }
