A string t is called nice if a string "2017" occurs in t as a subsequence but a string "2016" doesn‘t occur in t 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.
Limak has a string s of length n, with characters indexed 1 through n. He asks you q queries. In the i-th query you should compute and print the ugliness of a substring (continuous subsequence) of s starting at the index ai and ending at the index bi (inclusive).
The first line of the input contains two integers n and q (4 ≤ n ≤ 200 000, 1 ≤ q ≤ 200 000) — the length of the string s and the number of queries respectively.
The second line contains a string s of length n. Every character is one of digits ‘0‘–‘9‘.
The i-th of next q lines contains two integers ai and bi (1 ≤ ai ≤ bi ≤ n), describing a substring in the i-th query.
For each query print the ugliness of the given substring.
8 3
20166766
1 8
1 7
2 8
4
3
-1
15 5
012016662091670
3 4
1 14
4 15
1 13
10 15
-1
2
1
-1
-1
4 2
1234
2 4
1 2
-1
-1
In the first sample:
- In the first query, ugliness("20166766") = 4 because all four sixes must be removed.
- In the second query, ugliness("2016676") = 3 because all three sixes must be removed.
- In the third query, ugliness("0166766") = - 1 because it‘s impossible to remove some digits to get a nice string.
In the second sample:
- In the second query, ugliness("01201666209167") = 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("016662091670") = 1. It‘s optimal to remove the last digit ‘6‘, what gives a nice string "01666209170".
题目大意
给定一个长度为$n$的数字串。每次询问一个区间至少要删除多少个数字使得包含子序列"2017"但不包含子序列"2016"，无解输出-1。
dp是显然的。
因为每次询问一个区间，所以需要把dp状态扔到某个数据结构上。先考虑线段树。
线段树更新的时候是拿两段的信息合并，所以不能像做1~n的dp那样记录状态。
考虑2017之间的间隔：
| 2 | 0 | 1 | 7 |
0 1 2 3 4
线段树的每个节点存一个矩阵$A$。$a_{ij}$表示使原串的子序列包含2017中第$i$个间隔到第$j$个间隔组成的子串，但不包含严格包含它的子序列最少需要删除的数字、
转移是显然的，和区间dp一样。枚举区间，枚举中间点，然后转移就好了。
考虑初值问题，显然的是非2、0、1、7、6的数字对答案不影响，所以令$a_{ii} = 0$，$a_{ij} = \infty \ \ \ \left ( i \neq j \right )$。
考虑当前数字是2的时候，如果我希望只包含子串$[0, 0]$（这里表示两个间隔间的子串），那么就必须删掉这个2，故a_{00} = 1，如果希望包含子串$[0, 1]$，那么什么都不用做，所以$a_{01} = 0$。对于0、1、7同理。
考虑当前数字是6的时候，那么遇到子串$[i, 3]$希望转移回自己，那么需要付出1的代价，因为否则会包含子序列"2016"，同样如果遇到子串$[i, 4]$希望转移回自己，那么也需要付出1的代价。
由于很早以前过的这道题，所以不想重写一份，代码有点丑，请谅解。
Code
1 /** 2 * Codeforces 3 * Problem#750E 4 * Accepted 5 * Time: 998ms 6 * Memory: 49276k 7 */ 8 #include<iostream> 9 #include<fstream> 10 #include<sstream> 11 #include<cstdio> 12 #include<cstdlib> 13 #include<cstring> 14 #include<ctime> 15 #include<cctype> 16 #include<cmath> 17 #include<algorithm> 18 #include<stack> 19 #include<queue> 20 #include<set> 21 #include<map> 22 #include<vector> 23 #include<malloc.h> 24 #ifndef WIN32 25 #define AUTO "%lld" 26 #else 27 #define AUTO "%I64d" 28 #endif 29 using namespace std; 30 typedef bool boolean; 31 #define inf 0xfffffff 32 #define smin(a, b) (a) = min((a), (b)) 33 #define smax(a, b) (a) = max((a), (b)) 34 template<typename T> 35 inline void readInteger(T& u){ 36 char x; 37 int aFlag = 1; 38 while(!isdigit((x = getchar())) && x != ‘-‘ && x != -1); 39 if(x == -1) return; 40 if(x == ‘-‘){ 41 x = getchar(); 42 aFlag = -1; 43 } 44 for(u = x - ‘0‘; isdigit((x = getchar())); u = (u << 3) + (u << 1) + x - ‘0‘); 45 ungetc(x, stdin); 46 u *= aFlag; 47 } 48 49 typedef class Matrix { 50 public: 51 int mat[5][5]; //0: 1:2 2:20 3:201 4:2017 52 Matrix(){ } 53 Matrix(char x){ 54 for(int i = 0; i <= 4; i++) 55 for(int j = 0; j <= 4; j++) 56 mat[i][j] = (i == j) ? (0) : (inf); 57 if(x == ‘2‘) mat[0][0] = 1, mat[0][1] = 0; 58 else if(x == ‘0‘) mat[1][1] = 1, mat[1][2] = 0; 59 else if(x == ‘1‘) mat[2][2] = 1, mat[2][3] = 0; 60 else if(x == ‘7‘) mat[3][3] = 1, mat[3][4] = 0; 61 else if(x == ‘6‘) mat[3][3] = 1, mat[4][4] = 1; 62 } 63 64 Matrix operator +(Matrix x) { 65 Matrix res; 66 for(int i = 0; i < 5; i++) 67 for(int j = 0; j < 5; j++){ 68 res.mat[i][j] = inf; 69 for(int k = 0; k < 5; k++) 70 smin(res.mat[i][j], mat[i][k] + x.mat[k][j]); 71 } 72 return res; 73 } 74 }Matrix; 75 76 typedef class SegTreeNode { 77 public: 78 Matrix a; 79 SegTreeNode* left, *right; 80 SegTreeNode():left(NULL), right(NULL){ } 81 SegTreeNode(char x):left(NULL), right(NULL){ 82 a = Matrix(x); 83 } 84 85 void pushUp() { 86 a = left->a + right->a; 87 } 88 }SegTreeNode; 89 90 typedef class SegTree { 91 public: 92 SegTreeNode* root; 93 SegTree():root(NULL){ } 94 SegTree(int size, char* str){ 95 build(root, 1, size, str); 96 } 97 98 void build(SegTreeNode*& node, int l, int r, char* list){ 99 if(l == r){ 100 node = new SegTreeNode(list[l]); 101 return; 102 } 103 node = new SegTreeNode(); 104 int mid = (l + r) >> 1; 105 build(node->left, l, mid, list); 106 build(node->right, mid + 1, r, list); 107 node->pushUp(); 108 } 109 110 Matrix query(SegTreeNode*& node, int l, int r, int from, int end) { 111 if(l == from && r == end) return node->a; 112 int mid = (l + r) >> 1; 113 if(end <= mid) return query(node->left, l, mid, from, end); 114 if(from > mid) return query(node->right, mid + 1, r, from, end); 115 return query(node->left, l, mid, from, mid) + query(node->right, mid + 1, r, mid + 1, end); 116 } 117 }SegTree; 118 119 int n, q; 120 char* str; 121 SegTree st; 122 123 inline void init() { 124 readInteger(n); 125 readInteger(q); 126 str = new char[(const int)(n + 5)]; 127 scanf("%s", str + 1); 128 st = SegTree(n, str); 129 } 130 131 inline void solve() { 132 int a, b; 133 while(q--) { 134 readInteger(a); 135 readInteger(b); 136 Matrix c = st.query(st.root, 1, n, a, b); 137 printf("%d\n", (c.mat[0][4] == inf) ? (-1) : (c.mat[0][4])); 138 } 139 } 140 141 int main() { 142 init(); 143 solve(); 144 return 0; 145 }