题目链接
题目大意:
我们定义合法的字符串为 T, (T), ((T)(T))… 他给出n个操作,RL代表左移,剩下的字符就表示把当前位置上的字符替换为现在的字符。现在有个染色规则,如果两个括号是内嵌(nested)的关系,那他们的颜色就一定要不一样,然后输出每步操作后染色的最少颜色,如果不是合法字符串, 那就输出-1.
仔细一想,最少颜色其实就是字符串的最大深度。我们可以用线段树维护1~1e6区间内的区间和的最大值和最小值,如果区间和不为0或者最小值 < 0,那就不是合法字符串,输出-1
ac代码
#include <bits/stdc++.h>
#define mp make_pair
#define mt make_tuple
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define forn(i, n) for (int i = 0; i < (int)(n); ++i)
#define for1(i, n) for (int i = 1; i <= (int)(n); ++i)
#define ford(i, n) for (int i = (int)(n) - 1; i >= 0; --i)
#define fore(i, a, b) for (int i = (int)(a); i <= (int)(b); ++i)
#define rep(i, l, r) for (int i = (l); i <= (r); i++)
#define per(i, r, l) for (int i = (r); i >= (l); i--)
#define lson node<<1
#define rson node<<1|1
using namespace std;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<pii> vpi;
typedef vector<vi> vvi;
typedef long long i64;
typedef vector<i64> vi64;
typedef vector<vi64> vvi64;
typedef pair<i64, i64> pi64;
typedef double ld;
template<class T> bool uin(T &a, T b) { return a > b ? (a = b, true) : false; }
template<class T> bool uax(T &a, T b) { return a < b ? (a = b, true) : false; }
const int maxn = (int)1e6 + 1000;
const int N = (int)1e6 + 100;
int n, sum[maxn << 2], mi[maxn << 2], ma[maxn << 2];
string s;
char x[maxn];
void build(int l, int r, int node) {
sum[node] = mi[node] = ma[node] = 0;
if (l == r) return;
int mid = (l + r) >> 1;
build(l, mid, lson);
build(mid + 1, r, rson);
}
void pushup(int node) {
sum[node] = sum[lson] + sum[rson];
mi[node] = min(mi[lson], sum[lson] + mi[rson]);
ma[node] = max(ma[lson], sum[lson] + ma[rson]);
}
void update(int l, int r, int node, int pos, int num) {
if (l == r) {
sum[node] = mi[node] = ma[node] = num;
return;
}
int mid = (l + r) >> 1;
if (pos <= mid) update(l, mid, lson, pos, num);
else update(mid + 1, r, rson, pos, num);
pushup(node);
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
cout.precision(10);
cout << fixed;
#ifdef LOCAL_DEFINE
freopen("in", "r", stdin);
#endif
cin >> n;
cin >> s;
int now = 1;
build(1, N, 1);
forn(i, n) {
if (s[i] == 'R') ++now;
else if (s[i] == 'L') {
if (now > 1) --now;
} else if (s[i] == '(') {
update(1, N, 1, now, 1);
x[now] = '(';
} else if (s[i] == ')') {
update(1, N, 1, now, -1);
x[now] = ')';
} else {
char ch = x[now];
update(1, N, 1, now, 0);
}
if (sum[1] != 0 || mi[1] < 0) cout << -1 << " ";
else cout << ma[1] << " ";
}
#ifdef LOCAL_DEFINE
cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
#endif
return 0;
}