【codeforces 750E】New Year and Old Subsequence

time limit per test3 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
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 3
20166766
1 8
1 7
2 8
output
4
3
-1
input
15 5
012016662091670
3 4
1 14
4 15
1 13
10 15
output
-1
2
1
-1
-1
input
4 2
1234
2 4
1 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”.

【题目链接】:http://codeforces.com/contest/750/problem/E

【题解】

    merge矩阵(dp)+线段树;    m[i][j]表示从状态i->j转移的花费;    0表示什么都没有    1表示出现了2    2表示出现了20    3表示出现了201    4表示出现了2017    出现了2,20，则遇到6的时候我么可以不用管;    但是如果出现了201或2017,然后又遇到了6,则我们需要把这个6删掉;     for (i = 0;i <= 4;i++)//除了遇到2,0,1,6,7这5个特别的数字之外,        m[i][i] = 0;//遇到的时候状态都不会变;所以不用花费(操作);    if (ch == '2')    {        m[0][0] = 1;//什么都没有->什么都没有,删掉这个2        m[0][1] = 0;//什么都没有->出现了2,花费变成0        //注意这里m[1][1] = 0,表示从出现了一个2到出现了一个2可以不用删掉任何东西,因为加上一个2也无妨,下面的解释就省略了,希望大家能看得明白.    }    if (ch=='0')    {        m[1][1] = 1 ;//出现了一个2->出现了一个2,则把0删掉        m[1][2] = 0;//从出现一个2->出现了20,因为当前就是0,所以什么都不用加    }    if (ch=='1')    {        ma[2][2] = 1;//出现了20->出现了20,则把1删掉        ma[2][3] = 0;//出现了20->出现了201，因为刚好遇到一个1则什么都不加    }    if (ch=='7')    {        m[3][3] = 1;//出现了201->出现了201,则把遇到的7删掉        m[3][4] = 0;//出现了201->出现了2017,刚好遇到7，则什么都不做    }    if (ch=='6')    {        a[3][3] = 1;//出现了201->出现了数字6,则一定要把6删掉,不然无法满足题意        a[4][4] = 1;//出现了2017->出现了数字6,则也一定要把6删掉.    }    线段树加一个合并操作就好;    那个合并的操作和floyd算法类似;    最后输出m[0][4];表示从什么都没有然后出现"2017"

【完整代码】

#include <bits/stdc++.h>using namespace std;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define LL long long#define rep1(i,a,b) for (int i = a;i <= b;i++)#define rep2(i,a,b) for (int i = a;i >= b;i--)#define mp make_pair#define pb push_back#define fi first#define se second#define rei(x) scanf("%d",&x)#define rel(x) scanf("%I64d",&x)typedef pair<int,int> pii;typedef pair<LL,LL> pll;const int MAXN = 2e5+10;const int INF = 7e8;const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};const double pi = acos(-1.0);struct abc{    int m[5][5];    friend abc operator * (abc a,abc b)    {        abc d;        rep1(i,0,4)            rep1(j,0,4)            {                d.m[i][j] = INF;                rep1(k,0,4)                    d.m[i][j] = min(d.m[i][j],a.m[i][k]+b.m[k][j]);            }        return d;    }};int n,q;abc a[MAXN*4];char s[MAXN];void build(int l,int r,int rt){    if (l==r)    {        char ch = s[l];        rep1(i,0,4)            rep1(j,0,4)                a[rt].m[i][j] = (i==j)?0:INF;        if (ch == '2')        {            a[rt].m[0][0] = 1;            a[rt].m[0][1] = 0;        }        if (ch=='0')        {            a[rt].m[1][1] = 1;            a[rt].m[1][2] = 0;        }        if (ch=='1')        {            a[rt].m[2][2] = 1;            a[rt].m[2][3] = 0;        }        if (ch=='7')        {            a[rt].m[3][3] = 1;            a[rt].m[3][4] = 0;        }        if (ch=='6')        {            a[rt].m[3][3] = 1;            a[rt].m[4][4] = 1;        }        return;    }    int m = (l+r)>>1;    build(lson);build(rson);    a[rt] = a[rt<<1]*a[rt<<1|1];}abc query(int L,int R,int l,int r,int rt){    if (L<=l && r <= R)        return a[rt];    int m = (l+r)>>1;    abc temp1,temp2;    bool f1 = 0,f2 = 0;    if (L <= m)    {        temp1 = query(L,R,lson);        f1 = 1;    }    if (m < R)    {        temp2 = query(L,R,rson);        f2 = 1;    }    if (f1&&f2)        return temp1*temp2;    if (f1)        return temp1;    if (f2)        return temp2;}int main(){    //freopen("F:\\rush.txt","r",stdin);    rei(n);rei(q);    scanf("%s",s+1);    build(1,n,1);    rep1(i,1,q)    {        int L,R;        rei(L);rei(R);        int ans = query(L,R,1,n,1).m[0][4];        if (ans>=INF)            puts("-1");        else            cout << ans << endl;    }    return 0;}