UESTC 1546 bracket sequence (segment update)

Question: There are T (t <10) groups of test data, then there is a number n (n <100000), and the next line contains n characters, each character is '(' or ')'. There are three operations (1) "set l r c", which indicates changing all elements in the range [L, R] to C, C is one of '(' or. (2) "Reverse l r", indicating that the '(' and ')' in the range [L, R] is reversed. (3) "query l r" indicates whether the query interval [L, R] is a legal sequence of parentheses.

If we regard '(' As-1, and ')' as 1, note that if we start from L in the range [L, R] and continue to backward until R and sum at the same time, in this process, if the maximum value of the sum is less than or equal to 0, and the final sum is equal to or equal to 0, the sequence is legal, that is, the output is "yes ".

This requires that a sum domain and MX domain (that is, the maximum value) be saved in the node of the online segment tree ), there are two delay operations corresponding to the operation "set" and the operation "reverse", namely flag_set and flag_rev. Because the operation "reverse" requires switching between 1 and-1, in order to maintain the MX domain, a domain is added to the node of the online segment tree, mi (indicating the minimum value ), in this way, you can obtain the MX in this range through some operations when performing the "reverse" operation.

To pass down the delayed operation, you must first pass the "set" Operation mark, namely, flag_set, and then pass flag_rev. Because when you perform the "set" operation on a range, the mark of the previous "reverse" operation is set to 0 (because the previous "reverse" operation is invalid ), if a range contains both flag_set and flag_rev, the "set" operation first reaches this range, and then the "reverse" Operation reaches this range.

When the interval is merged, the MI domain of the current interval is from. The MI domain of the Left son and the sum of the Left son plus the MI of the right son take the minimum value, and the update of the MX domain is similar.

#include <iostream>#include <cstdio>#include <cstring>using namespace std;#define LL(x) (x<<1)#define RR(x) (x<<1|1)#define MID(a,b) (a+((b-a)>>1))const int N=100005;struct node{int lft,rht;int mi,mx,sum;int flag_set,flag_rev;int mid(){return MID(lft,rht);}int len(){return rht-lft+1;}void init(){ flag_set=flag_rev=mi=mx=sum=0; }void fun(int valu){if(valu){mi=min(valu,valu*len());mx=max(valu,valu*len());sum=valu*len();flag_set=valu; flag_rev=0;}else{sum=-sum; swap(mi,mx); mi=-mi; mx=-mx;flag_rev^=1;}}};int y[N],n,m;struct Segtree{node tree[N*4];void down(int ind){if(tree[ind].flag_set){tree[LL(ind)].fun(tree[ind].flag_set);tree[RR(ind)].fun(tree[ind].flag_set);tree[ind].flag_set=0;}if(tree[ind].flag_rev){tree[LL(ind)].fun(0);tree[RR(ind)].fun(0);tree[ind].flag_rev=0;}}void up(int ind){tree[ind].sum=tree[LL(ind)].sum+tree[RR(ind)].sum;tree[ind].mx=max(tree[LL(ind)].mx,tree[LL(ind)].sum+tree[RR(ind)].mx);tree[ind].mi=min(tree[LL(ind)].mi,tree[LL(ind)].sum+tree[RR(ind)].mi);}void build(int lft,int rht,int ind){tree[ind].lft=lft;tree[ind].rht=rht;tree[ind].init();if(lft==rht) tree[ind].mi=tree[ind].mx=tree[ind].sum=y[lft];else {int mid=tree[ind].mid();build(lft,mid,LL(ind));build(mid+1,rht,RR(ind));up(ind);}}void updata(int st,int ed,int ind,int valu){int lft=tree[ind].lft,rht=tree[ind].rht;if(st<=lft&&rht<=ed) tree[ind].fun(valu);else {down(ind);int mid=tree[ind].mid();if(st<=mid) updata(st,ed,LL(ind),valu);if(ed> mid) updata(st,ed,RR(ind),valu);up(ind);}}void query(int st,int ed,int ind,int &mx,int &sum){int lft=tree[ind].lft,rht=tree[ind].rht;if(st<=lft&&rht<=ed){mx=tree[ind].mx;sum=tree[ind].sum;}else {down(ind);int mid=tree[ind].mid();if(ed<=mid) query(st,ed,LL(ind),mx,sum);else if(st>mid) query(st,ed,RR(ind),mx,sum);else {int mx1,mx2,sum1,sum2;query(st,ed,LL(ind),mx1,sum1);query(st,ed,RR(ind),mx2,sum2);mx=max(mx1,sum1+mx2); sum=sum1+sum2;}up(ind);}}}seg;int main(){int t,t_cnt=0;scanf("%d",&t);while(t--){char str[N];scanf("%d%s",&n,str);for(int i=0;i<n;i++){if(str[i]=='(') y[i]=-1;else y[i]=1;}seg.build(0,n-1,1);scanf("%d",&m);printf("Case %d:\n",++t_cnt);while(m--){int a,b;char cmd[10],chr[10];scanf("%s",cmd);if(cmd[0]=='s'){scanf("%d%d%s",&a,&b,chr);seg.updata(a,b,1,(chr[0]=='('?-1:1));}else if(cmd[0]=='r'){scanf("%d%d",&a,&b);seg.updata(a,b,1,0);}else{int mx,sum;scanf("%d%d",&a,&b);seg.query(a,b,1,mx,sum);if(mx<=0&&sum==0) puts("YES");else puts("NO");}}puts("");}return 0;}