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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 }
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原文地址：http://www.cnblogs.com/jklongint/p/4489298.html
