您的位置:首页 > 其它

洛谷 红魔馆OI 妖梦斩木棒

2017-07-08 21:03 363 查看
题目传送门

一眼看出线段树,记录区间方案数,左/右非x木棒数就可以了。

tips:不知道分块可否?

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cctype>
#define ll long long
#define INF 1000000000
#define clr(x) memset(x,0,sizeof(x))
using namespace std;

inline int read()
{
register int ret=0,c=getchar();
while(!isdigit(c))c=getchar();
while(isdigit(c))ret=ret*10+c-'0',c=getchar();
return ret;
}

#define M 200005

struct tree2
{
tree2 *lson,*rson;
int ln,rn,n;
}dizhi[15+M<<1],*t=&dizhi[0],*T,*root=++t;

int n,m;

void pushup(tree2 *tree,tree2 *lson,tree2 *rson)
{
int mer=min(lson->rn,rson->ln);
tree->n=lson->n+rson->n+(mer>0);
if(!mer&&!lson->n)tree->ln=max(lson->ln,rson->ln);
else tree->ln=lson->ln;
if(!mer&&!rson->n)tree->rn=max(lson->rn,rson->rn);
else tree->rn=rson->rn;
}

void bulid(tree2 *tree,int l,int r)
{
if(l==r)
{
if(l==1)tree->rn=1;
else if(l==n)tree->ln=1;
return ;
}
int mid=(l+r)>>1;
tree->lson=++t;
tree->rson=++t;
bulid(tree->lson,l,mid);
bulid(tree->rson,mid+1,r);
pushup(tree,tree->lson,tree->rson);
}

void change(tree2 *tree,int l,int r,int x,int num)
{
if(l==r)
{
tree->ln=tree->rn=0;
if(num==2)tree->ln=1;//')'
if(num==3)tree->rn=1;//'('
return ;
}
int mid=(l+r)>>1;
if(x<=mid)change(tree->lson,l,mid,x,num);
else change(tree->rson,mid+1,r,x,num);
pushup(tree,tree->lson,tree->rson);
}

tree2 *query(tree2 *tree,int l,int r,int x,int y)
{
if(x<=l&&y>=r)return tree;
int mid=(l+r)>>1;
tree2 *t1=NULL,*t2=NULL;
if(x<=mid)t1=query(tree->lson,l,mid,x,y);
if(y>mid)t2=query(tree->rson,mid+1,r,x,y);
if(t1==NULL)return t2;
if(t2==NULL)return t1;
tree2 *ret=++t;
pushup(ret,t1,t2);
return ret;
}

inline int get_num(char c)
{
if(c=='X')return 1;
if(c==')')return 2;
if(c=='(')return 3;
}

int main()
{
n=read(),m=read();
bulid(root,1,n);
T=t;
for(int i=1;i<=m;i++)
{
t=T+1;
int x=read();
if(x==1)
{
int a=read(),b=get_num(getchar());
change(root,1,n,a,b);
}
if(x==2)
{
int a=read(),b=read();
printf("%d\n",query(root,1,n,a,b)->n);
}
}
return 0;
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: