Codeforces 750 E New Year and Old Subsequence
Posted widsom
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了Codeforces 750 E New Year and Old Subsequence相关的知识,希望对你有一定的参考价值。
E. New Year and Old Subsequence
思路:线段树维护矩阵乘法。
代码:
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define y1 y11
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb emplace_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pli pair<LL, int>
#define pii pair<int, int>
#define piii pair<pii, int>
#define pdd pair<double, double>
#define mem(a, b) memset(a, b, sizeof(a))
#define debug(x) cerr << #x << " = " << x << "
";
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
//head
const int INF = 0x3f3f3f3f;
const int N = 2e5 + 10;
struct Matrix {
int a[5][5];
inline void init() {
memset(a, INF, sizeof a);
}
inline Matrix operator * (const Matrix & rhs) const {
Matrix res;
res.init();
for (int k = 0; k < 5; ++k) {
for (int i = 0; i < 5; ++i) {
for (int j = 0; j < 5; ++j) {
res.a[i][j] = min(res.a[i][j], a[i][k]+rhs.a[k][j]);
}
}
}
return res;
}
}tree[N<<2];
char s[N];
int n, q, l, r;
inline int push_up(int rt) {
tree[rt] = tree[rt<<1]*tree[rt<<1|1];
}
/* |2|0|1|7| */
void build(int rt, int l, int r) {
if(l == r) {
tree[rt].init();
for (int i = 0; i < 5; ++i) tree[rt].a[i][i] = 0;
if(s[l] == '2') tree[rt].a[0][1] = 0, tree[rt].a[0][0] = 1;
else if(s[l] == '0') tree[rt].a[1][2] = 0, tree[rt].a[1][1] = 1;
else if(s[l] == '1') tree[rt].a[2][3] = 0, tree[rt].a[2][2] = 1;
else if(s[l] == '7') tree[rt].a[3][4] = 0, tree[rt].a[3][3] = 1;
else if(s[l] == '6') tree[rt].a[3][3] = 1, tree[rt].a[4][4] = 1;
return ;
}
int m = l+r >> 1;
build(ls);
build(rs);
push_up(rt);
}
Matrix query(int L, int R, int rt, int l, int r) {
if(L <= l && r <= R) return tree[rt];
int m = l+r >> 1;
Matrix res;
res.init();
for (int i = 0; i < 5; ++i) res.a[i][i] = 0;
if(L <= m) res = res * query(L, R, ls);
if(R > m) res = res* query(L, R, rs);
return res;
}
int main() {
scanf("%d %d", &n, &q);
scanf("%s", s+1);
build(1, 1, n);
while(q--) {
scanf("%d %d", &l, &r);
Matrix res = query(l, r, 1, 1, n);
if(res.a[0][4] == INF) printf("-1
");
else printf("%d
", res.a[0][4]);
}
return 0;
}
以上是关于Codeforces 750 E New Year and Old Subsequence的主要内容,如果未能解决你的问题,请参考以下文章
[Codeforces 750E]New Year and Old Subsequence
CodeForces - 750D New Year and Fireworks
codeforces 750D New Year and FireworksDFS
Codeforces 750E New Year and Old Subsequence - 线段树 - 动态规划