题意：给定一个只含‘(‘,‘)‘的括号序列，有m个操作，改变某个位置的括号或者询问区间[L,R]内的括号进行配对后剩下的第K个括号的位置（配对的括号从原序列中删掉）。
思路：首先对于一个括号序列，进行配对后剩下的括号序列必定是")))...)((...(("这种形式的，令(x,y)表示当前区间的剩下的右括号数为x，左括号数为y这个状态，不难发现它可以通过子区间来合并，因此考虑用线段树解决。对于单点更新比较容易，直接在叶子节点改变一下，然后向上更新即可，对于L,R,K这个询问，首先进行一次区间[L,R]的查找，得到最终的左括号和右括号数目，如果数目和小于K，则答案为-1，否则可以确定第K个括号的是哪种类型的括号，假设第K个括号是右括号，则问题转化为求第K个右括号的位置，在线段树上二分即可，二分的时候需要注意，假定答案在右区间，且左区间有x个右括号y个左括号，则在右区间应查找第K-x+y个右括号。
1 #pragma comment(linker, "/STACK:102400000,102400000")
2
3 #include <iostream>
4 #include <cstdio>
5 #include <algorithm>
6 #include <cstdlib>
7 #include <cstring>
8 #include <map>
9 #include <queue>
10 #include <deque>
11 #include <cmath>
12 #include <ctime>
13 #include <cctype>
14 #include <set>
15 #include <bitset>
16 #include <functional>
17 #include <numeric>
18 #include <stdexcept>
19 #include <utility>
20 #include <vector>
21
22 using namespace std;
23
24 #define mem0(a) memset(a, 0, sizeof(a))
25 #define mem_1(a) memset(a, -1, sizeof(a))
26 #define lson l, m, rt << 1
27 #define rson m + 1, r, rt << 1 | 1
28 #define define_m int m = (l + r) >> 1
29 #define rep_up0(a, b) for (int a = 0; a < (b); a++)
30 #define rep_up1(a, b) for (int a = 1; a <= (b); a++)
31 #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--)
32 #define rep_down1(a, b) for (int a = b; a > 0; a--)
33 #define all(a) (a).begin(), (a).end()
34 #define lowbit(x) ((x) & (-(x)))
35 #define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {}
36 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
37 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
38 #define pchr(a) putchar(a)
39 #define pstr(a) printf("%s", a)
40 #define sstr(a) scanf("%s", a)
41 #define sint(a) scanf("%d", &a)
42 #define sint2(a, b) scanf("%d%d", &a, &b)
43 #define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c)
44 #define pint(a) printf("%d\n", a)
45 #define test_print1(a) cout << "var1 = " << a << endl
46 #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl
47 #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl
48
49 typedef long long LL;
50 typedef pair<int, int> pii;
51 typedef vector<int> vi;
52
53 const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1};
54 const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 };
55 const int maxn = 2e5 + 7;
56 const int md = 10007;
57 const int inf = 1e9 + 7;
58 const LL inf_L = 1e18 + 7;
59 const double pi = acos(-1.0);
60 const double eps = 1e-6;
61
62 template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);}
63 template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;}
64 template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;}
65 template<class T>T condition(bool f, T a, T b){return f?a:b;}
66 template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];}
67 int make_id(int x, int y, int n) { return x * n + y; }
68
69 int ans;
70 char s[maxn];
71
72 struct SegTree {
73 struct Node {
74 int cl, cr;
75 } tree[maxn << 2];
76
77 Node merge(Node b, Node c) {
78 Node a;
79 a.cl = c.cl;
80 a.cr = b.cr;
81 if (b.cl <= c.cr) a.cr += c.cr - b.cl;
82 else a.cl += b.cl - c.cr;
83 return a;
84 }
85
86 void build(int l, int r, int rt) {
87 if (l == r) {
88 char ch = s[l - 1];
89 tree[rt].cl = ch == ‘(‘;
90 tree[rt].cr = ch == ‘)‘;
91 return ;
92 }
93 define_m;
94 build(lson);
95 build(rson);
96 tree[rt] = merge(tree[rt << 1], tree[rt << 1 | 1]);
97 }
98
99 void update(int u, int l, int r, int rt) {
100 if (l == r) {
101 swap(tree[rt].cl, tree[rt].cr);
102 return ;
103 }
104 define_m;
105 if (u <= m) update(u, lson);
106 else update(u, rson);
107 tree[rt] = merge(tree[rt << 1], tree[rt << 1 | 1]);
108 }
109
110 Node query(int L, int R, int l, int r, int rt) {
111 if (L <= l && r <= R) return tree[rt];
112 define_m;
113 if (R <= m) return query(L, R, lson);
114 if (L > m) return query(L, R, rson);
115 return merge(query(L, R, lson), query(L, R, rson));
116 }
117
118 Node find1(int L, int R, int k, int l, int r, int rt) {
119 if (L <= l && r <= R && (k > tree[rt].cr || k <= 0 || ans)) return tree[rt];
120 if (l == r) {
121 ans = l;
122 return tree[rt];
123 }
124 define_m;
125 if (R <= m) return find1(L, R, k, lson);
126 if (L > m) return find1(L, R, k, rson);
127 Node buf = find1(L, R, k, lson);
128 if (ans) return buf;
129 return merge(buf, find1(L, R, k - buf.cr + buf.cl, rson));
130 }
131
132 Node find2(int L, int R, int k, int l, int r, int rt) {
133 if (L <= l && r <= R && (k > tree[rt].cl || k <= 0 || ans)) return tree[rt];
134 if (l == r) {
135 ans = l;
136 return tree[rt];
137 }
138 define_m;
139 if (R <= m) return find2(L, R, k, lson);
140 if (L > m) return find2(L, R, k, rson);
141 Node buf = find2(L, R, k, rson);
142 if (ans) return buf;
143 return merge(find2(L, R, k - buf.cl + buf.cr, lson), buf);
144 }
145 };
146 SegTree ST;
147 int main() {
148 //freopen("in.txt", "r", stdin);
149 int T, n, m;
150 cin >> T;
151 while (T --) {
152 cin >> n >> m;
153 scanf("%s", s);
154 ST.build(1, n, 1);
155 rep_up0(i, m) {
156 int id;
157 sint(id);
158 if (id == 1) {
159 int u;
160 sint(u);
161 ST.update(u, 1, n, 1);
162 }
163 else {
164 int u, v, k;
165 sint3(u, v, k);
166 SegTree::Node buf = ST.query(u, v, 1, n, 1);
167 if (buf.cl + buf.cr < k) {
168 puts("-1");
169 continue;
170 }
171 ans = 0;
172 if (buf.cr >= k) ST.find1(u, v, k, 1, n, 1);
173 else ST.find2(u, v, buf.cl + buf.cr - k + 1, 1, n, 1);
174 printf("%d\n", ans);
175 }
176 }
177 }
178 return 0;
179 }
时间： 2024-10-09 23:41:58