CF750E New Year and Old Subsequence

讲道理好久没有做过题了..

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题目大意

给出长度为$n$的只含数字的串，有$q$个询问，每次询问一段区间，问最少删去多少个数才能变成只含2017子序列而不含2016子序列

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吉爸爸好强啊..

定义$a_{i,j}$表示该区间从第$i$位匹配不了第$j$位最少要删去的数字数

这个东西用一个线段树或者st表来维护，合并是类似于矩阵乘法的..

根据吉爸爸的优化，对于一个询问区间$[l,r]$

因为一定需要一个7，所以把它分成两段，就像这样

l...最后一个7...r

那么前面一段只需要求出从第一位但是匹配不了最后一位的方案数，再加上后面那一段6的个数就是答案了..

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code

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
using namespace std;
const int Maxn = 200010;
const int lg = 20;
const int inf = 0x7fffffff;
struct node {
	int a[4][4];
	void clear (){ memset ( a, 63, sizeof (a) ); }
}f[Maxn][lg], p;
int n, m;
char s[Maxn];
int bit[Maxn], pre[Maxn], sum[Maxn];
int _min ( int x, int y ){ return x < y ? x : y; }
void merge ( node &ret, node x, node y ){
	ret.clear ();
	int i, j, k;
	for ( i = 0; i < 4; i ++ ) for ( j = 0; j < 4; j ++ ) for ( k = 0; k < 4; k ++ ) ret.a[i][j] = _min ( x.a[i][k]+y.a[k][j], ret.a[i][j] );
}
int getans ( int l, int r ){
	int len = r-l+1;
	p.clear ();
	p.a[0][0] = p.a[1][1] = p.a[2][2] = p.a[3][3] = 0;
	while ( len > 0 ){
		merge ( p, p, f[l][bit[len]] );
		l += (1<<bit[len]);
		len -= (1<<bit[len]);
	}
	return p.a[0][3];
}
int main (){
	int i, j, k;
	scanf ( "%d%d", &n, &m );
	scanf ( "%s", s+1 );
	bit[1] = 0;
	for ( i = 2; i <= n; i ++ ) bit[i] = bit[i>>1]+1;
	for ( i = 1; i <= n; i ++ ){
		if ( s[i] == '6' ) sum[i] = sum[i-1]+1; else sum[i] = sum[i-1];
		if ( s[i] == '7' ) pre[i] = i; else pre[i] = pre[i-1];
	}
	for ( i = 1; i <= n; i ++ ){
		f[i][0].clear ();
		f[i][0].a[0][0] = f[i][0].a[1][1] = f[i][0].a[2][2] = f[i][0].a[3][3] = 0;
		if ( s[i] == '2' ) f[i][0].a[0][1] = 0, f[i][0].a[0][0] = 1;
		if ( s[i] == '0' ) f[i][0].a[1][2] = 0, f[i][0].a[1][1] = 1;
		if ( s[i] == '1' ) f[i][0].a[2][3] = 0, f[i][0].a[2][2] = 1;
		if ( s[i] == '6' ) f[i][0].a[3][3] = 1;
	}
	for ( i = n; i >= 1; i -- ){
		for ( j = 1; j <= 18 && i+(1<<(j-1)) <= n; j ++ ){
			merge ( f[i][j], f[i][j-1], f[i+(1<<(j-1))][j-1] );
		}
	}
	for ( i = 1; i <= m; i ++ ){
		int l, r;
		scanf ( "%d%d", &l, &r );
		k = pre[r];
		int ans = getans ( l, k-1 );
		if ( ans > n ) printf ( "-1\n" );
		else printf ( "%d\n", ans+sum[r]-sum[k] );
	}
	return 0;
}

哎好久没打都有点生疏了.. 加油吧..