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).
Input
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.
Output
For each query print the ugliness of the given substring.
Examples
input
8 3201667661 81 72 8
output
43-1
input
15 50120166620916703 41 144 151 1310 15
output
-121-1-1
input
4 212342 41 2
output
-1-1
Note
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 }
原文地址：https://www.cnblogs.com/yyf0309/p/8475433.html