[hdu5217]线段树

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。

 #pragma comment(linker, "/STACK:102400000,102400000")

 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <cstdlib>
 #include <cstring>
 #include <map>
 #include <queue>
 #include <deque>
 #include <cmath>
 #include <ctime>
 #include <cctype>
 #include <set>
 #include <bitset>
 #include <functional>
 #include <numeric>
 #include <stdexcept>
 #include <utility>
 #include <vector>

 using namespace std;

 #define mem0(a) memset(a, 0, sizeof(a))
 #define mem_1(a) memset(a, -1, sizeof(a))
 #define lson l, m, rt << 1
 #define rson m + 1, r, rt << 1 | 1
 #define define_m int m = (l + r) >> 1
 #define rep_up0(a, b) for (int a = 0; a < (b); a++)
 #define rep_up1(a, b) for (int a = 1; a <= (b); a++)
 #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--)
 #define rep_down1(a, b) for (int a = b; a > 0; a--)
 #define all(a) (a).begin(), (a).end()
 #define lowbit(x) ((x) & (-(x)))
 #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) {}
 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
 #define pchr(a) putchar(a)
 #define pstr(a) printf("%s", a)
 #define sstr(a) scanf("%s", a)
 #define sint(a) scanf("%d", &a)
 #define sint2(a, b) scanf("%d%d", &a, &b)
 #define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c)
 #define pint(a) printf("%d\n", a)
 #define test_print1(a) cout << "var1 = " << a << endl
 #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl
 #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl

 typedef long long LL;
 typedef pair<int, int> pii;
 typedef vector<int> vi;

 const int dx[] = {, , -, , , , -, -};
 const int dy[] = {-, , , , , -, , - };
 const int maxn = 2e5 + ;
 const int md = ;
 const int inf = 1e9 + ;
 const LL inf_L = 1e18 + ;
 const double pi = acos(-1.0);
 const double eps = 1e-;

 template<class T>T gcd(T a, T b){return b==?a:gcd(b,a%b);}
 template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;}
 template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;}
 template<class T>T condition(bool f, T a, T b){return f?a:b;}
 template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];}
 int make_id(int x, int y, int n) { return x * n + y; }

 int ans;
 char s[maxn];

 struct SegTree {
     struct Node {
         int cl, cr;
     } tree[maxn << ];

     Node merge(Node b, Node c) {
         Node a;
         a.cl = c.cl;
         a.cr = b.cr;
         if (b.cl <= c.cr) a.cr += c.cr - b.cl;
         else a.cl += b.cl - c.cr;
         return a;
     }

     void build(int l, int r, int rt) {
         if (l == r) {
             char ch = s[l - ];
             tree[rt].cl = ch == '(';
             tree[rt].cr = ch == ')';
             return ;
         }
         define_m;
         build(lson);
         build(rson);
         tree[rt] = merge(tree[rt << ], tree[rt <<  | ]);
     }

     void update(int u, int l, int r, int rt) {
         if (l == r) {
             swap(tree[rt].cl, tree[rt].cr);
             return ;
         }
         define_m;
         if (u <= m) update(u, lson);
         else update(u, rson);
         tree[rt] = merge(tree[rt << ], tree[rt <<  | ]);
     }

     Node query(int L, int R, int l, int r, int rt) {
         if (L <= l && r <= R)  return tree[rt];
         define_m;
         if (R <= m) return query(L, R, lson);
         if (L > m) return query(L, R, rson);
         return merge(query(L, R, lson), query(L, R, rson));
     }

     Node find1(int L, int R, int k, int l, int r, int rt) {
         if (L <= l && r <= R && (k > tree[rt].cr || k <=  || ans)) return tree[rt];
         if (l == r) {
             ans = l;
             return tree[rt];
         }
         define_m;
         if (R <= m) return find1(L, R, k, lson);
         if (L > m) return find1(L, R, k, rson);
         Node buf = find1(L, R, k, lson);
         if (ans) return buf;
         return merge(buf, find1(L, R, k - buf.cr + buf.cl, rson));
     }

      Node find2(int L, int R, int k, int l, int r, int rt) {
         if (L <= l && r <= R && (k > tree[rt].cl || k <=  || ans)) return tree[rt];
         if (l == r) {
             ans = l;
             return tree[rt];
         }
         define_m;
         if (R <= m) return find2(L, R, k, lson);
         if (L > m) return find2(L, R, k, rson);
         Node buf = find2(L, R, k, rson);
         if (ans) return buf;
         return merge(find2(L, R, k - buf.cl + buf.cr, lson), buf);
     }
 };
 SegTree ST;
 int main() {
     //freopen("in.txt", "r", stdin);
     int T, n, m;
     cin >> T;
     while (T --) {
         cin >> n >> m;
         scanf("%s", s);
         ST.build(, n, );
         rep_up0(i, m) {
             int id;
             sint(id);
             if (id == ) {
                 int u;
                 sint(u);
                 ST.update(u, , n, );
             }
             else {
                 int u, v, k;
                 sint3(u, v, k);
                 SegTree::Node buf = ST.query(u, v, , n, );
                 if (buf.cl + buf.cr < k) {
                     puts("-1");
                     continue;
                 }
                 ans = ;
                 if (buf.cr >= k) ST.find1(u, v, k, , n, );
                 else ST.find2(u, v, buf.cl + buf.cr - k + , , n, );
                 printf("%d\n", ans);
             }
         }
     }
     return ;
 }