题目链接：https://codeforces.com/contest/1263/problem/E
题意：一个编译器，每次输入一些字符，R表示光标右移，L表示光标左移，然后有一些左括号（ 和 右括号 ），每次会询问当前输入的数据的括号是否合法，如果不合法输出-1，如果合法输出最大合法的括号对数，合法的括号就是（）这种形式的。
思路：大致思路是用线段树维护区间一个区间前缀和，初始化前缀和为0。遇到单点更新，（让管辖区间+1，）就让管辖区间-1，，判断是否是合法括号需要判断区间最小值是否为0，且保证1到n区间的前缀和为0（画图思考一下），满足两种条件的同时才能说明此为合法括号序列，最终查询区间最大值就是最大匹配的括号对数（画图思考）。
代码：
1 #include<iostream> 2 #include<cmath> 3 #include<algorithm> 4 #include<cstdio> 5 #include<vector> 6 #include<queue> 7 #include<cstring> 8 #include<set> 9 #define inf 0x3f3f3f3f 10 using namespace std; 11 typedef long long LL; 12 const int maxn = 1e6+10; 13 struct node{ 14 int l,r; 15 int Min,Max; 16 int val; 17 int lazy; 18 }tree[maxn*4+5]; 19 void pushdown(int index){ 20 if(tree[index].lazy){ 21 tree[index<<1].val += (tree[index<<1].r-tree[index<<1].l+1)*tree[index].lazy; 22 tree[index<<1|1].val +=(tree[index<<1|1].r-tree[index<<1|1].l+1)*tree[index].lazy; 23 tree[index<<1].Max += tree[index].lazy; 24 tree[index<<1|1].Max += tree[index].lazy; 25 tree[index<<1].Min += tree[index].lazy; 26 tree[index<<1|1].Min += tree[index].lazy; 27 tree[index<<1].lazy += tree[index].lazy; 28 tree[index<<1|1].lazy += tree[index].lazy; 29 tree[index].lazy = 0; 30 } 31 } 32 void pushup(int index){ 33 tree[index].val = tree[index<<1].val+tree[index<<1|1].val; 34 tree[index].Max = max(tree[index<<1].Max,tree[index<<1|1].Max); 35 tree[index].Min = min(tree[index<<1].Min,tree[index<<1|1].Min); 36 } 37 38 void build(int l,int r,int index){//建树 39 tree[index].l = l,tree[index].r = r,tree[index].lazy = 0; 40 if(l == r){ 41 //scanf("%d",&tree[index].val); 42 tree[index].val = 0; 43 tree[index].Max = tree[index].Min = tree[index].val; 44 return; 45 } 46 int mid = (l+r)>>1;// 47 build(l,mid,index<<1); 48 build(mid+1,r,index<<1|1); 49 pushup(index); 50 } 51 void update(int l,int r,int index,int val){//区间修改 52 if(l <= tree[index].l && r >= tree[index].r){ 53 tree[index].val += (tree[index].r-tree[index].l+1)*val; 54 tree[index].Max += val; 55 tree[index].Min += val; 56 tree[index].lazy += val;//延时标记 57 return ; 58 } 59 pushdown(index); 60 int mid = (tree[index].l+tree[index].r)>>1; 61 if(l <= mid){ 62 update(l,r,index<<1,val); 63 } 64 if(r > mid){ 65 update(l,r,index<<1|1,val); 66 } 67 pushup(index); 68 } 69 70 LL query_range(int l,int r,int index){//区间查询 71 if(l <= tree[index].l && r >= tree[index].r){ 72 return tree[index].val; 73 //return tree[index].mn; 74 } 75 pushdown(index); 76 int mid = (tree[index].l+tree[index].r)>>1; 77 LL res = 0; 78 LL Max = 0; 79 LL Min = inf; 80 if(l <= mid){ 81 res += query_range(l,r,index<<1); 82 } 83 if(r > mid){ 84 res += query_range(l,r,index<<1|1); 85 } 86 return res; 87 } 88 89 LL query_max(int l,int r,int index){//最大值查询 90 if(l<= tree[index].l && r>= tree[index].r ){ 91 return tree[index].Max; 92 } 93 pushdown(index); 94 int mid = (tree[index].l + tree[index].r )>>1; 95 LL ans = 0; 96 LL MAX = 0,MIN = inf; 97 if(l <= mid) MAX = max(query_max(l,r,index<<1),MAX); 98 if(r > mid) MAX = max(query_max(l,r,index<<1|1),MAX); 99 return MAX; 100 } 101 LL query_min(int l,int r,int index){//最小值查询 102 if(l<= tree[index].l && r>= tree[index].r ){ 103 return tree[index].Min; 104 } 105 pushdown(index); 106 int mid = (tree[index].l + tree[index].r )>>1; 107 LL ans = 0; 108 LL MAX = 0,MIN = inf; 109 if(l <= mid) MIN = min(query_min(l,r,index<<1),MIN); 110 if(r > mid) MIN = min(query_min(l,r,index<<1|1),MIN); 111 return MIN; 112 } 113 string s; 114 LL ss[maxn]; 115 int main(){ 116 int n; 117 scanf("%d",&n); 118 n = n+10; 119 build(1,n,1); 120 cin>>s; 121 int pos = 1; 122 vector<int> ans; 123 for(int i = 0;i<s.length() ;i++){ 124 if(s[i] == '('){ 125 update(pos,n,1,1-ss[pos]); 126 ss[pos] = 1; 127 } 128 else if(s[i] == ')'){ 129 update(pos,n,1,-1-ss[pos]); 130 ss[pos] = -1; 131 } 132 else if(s[i] == 'R'){ 133 pos++; 134 } 135 else if(s[i] == 'L'){ 136 if(pos>1) pos--; 137 } 138 else{ 139 if(ss[pos] == 1){ 140 update(pos,n,1,-1); 141 } 142 else if(ss[pos] == -1){ 143 update(pos,n,1,1); 144 } 145 ss[pos] = 0; 146 } 147 if(query_min(1,n,1) == 0 && query_range(n,n,1) == 0){ 148 ans.push_back(query_max(1,n,1)); 149 } 150 else{ 151 ans.push_back(-1); 152 } 153 } 154 for(int i = 0;i<ans.size() ;i++){ 155 cout<<ans[i]<<" "; 156 } 157 //cout<<endl; 158 return 0; 159 }