hdu 1540 Tunnel Warfare（线段树 连续区间）

Tunnel Warfare

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2069    Accepted Submission(s): 766

Problem Description

During the War of Resistance Against Japan, tunnel warfare was carried out extensively in the vast areas of north China Plain. Generally speaking, villages connected by tunnels lay in a line. Except the two at the ends, every village was directly connected with two neighboring ones.

Frequently the invaders launched attack on some of the villages and destroyed the parts of tunnels in them. The Eighth Route Army commanders requested the latest connection state of the tunnels and villages. If some villages are severely isolated, restoration of connection must be done immediately!

 

Input

The first line of the input contains two positive integers n and m (n, m ≤ 50,000) indicating the number of villages and events. Each of the next m lines describes an event.

There are three different events described in different format shown below:

D x: The x-th village was destroyed.

Q x: The Army commands requested the number of villages that x-th village was directly or indirectly connected with including itself.

R: The village destroyed last was rebuilt.

 

Output

Output the answer to each of the Army commanders’ request in order on a separate line.

 

Sample Input

   
   
   
   

    
    
    
    7 9
D 3
D 6
D 5
Q 4
Q 5
R
Q 4
R
Q 4
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    1
0
2
4
   
   
   
   

 

Source

POJ Monthly

 

Recommend

LL

 

题目：http://acm.hdu.edu.cn/showproblem.php?pid=1540

题意：有n个村庄，现在，每次摧毁一个村庄或修好一个村庄，或者问你和这个村庄连接的村庄个数。。。

分析：这题相比上题来讲，就是个超精简版吧，只要记录每个区间左边开始的连续村庄数，和右边开始的连续村庄数，每次摧毁或维修都是单点更新，而查找一个点，只要这个点，在 m-lmost[lson]+1~m+rmost[rson]之间，就说明它被这个区间的两个子区间包含了，直接返回就行，否则就是只被一个区间包含，继续递归。。。

代码：

#include<cstdio>
#include<iostream>
#define ls rt<<1
#define rs rt<<1|1
#define lson l,m,ls
#define rson m+1,r,rs
#define uprt rt,m-l+1,r-m
using namespace std;
const int mm=55555;
const int mn=mm<<2;
int lm[mn],rm[mn];
int q[mm];
void pushup(int rt,int l1,int l2)
{
    lm[rt]=lm[ls],rm[rt]=rm[rs];
    if(lm[rt]>=l1)lm[rt]+=lm[rs];
    if(rm[rt]>=l2)rm[rt]+=rm[ls];
}
void build(int l,int r,int rt)
{
    lm[rt]=rm[rt]=r-l+1;
    if(l==r)return;
    int m=(l+r)>>1;
    build(lson);
    build(rson);
}
void updata(int p,int x,int l,int r,int rt)
{
    if(l==r)
    {
        lm[rt]=rm[rt]=x;
        return;
    }
    int m=(l+r)>>1;
    if(p<=m)updata(p,x,lson);
    else updata(p,x,rson);
    pushup(uprt);
}
int query(int p,int l,int r,int rt)
{
    if(l==r)return 0;
    int m=(l+r)>>1;
    if(m-rm[ls]+1<=p&&m+lm[rs]>=p)return rm[ls]+lm[rs];
    if(p<=m)return query(p,lson);
    return query(p,rson);
}
int main()
{
    int i,n,m,top;
    char op[55];
    while(~scanf("%d%d",&n,&m))
    {
        build(1,n,1);
        top=0;
        while(m--)
        {
            scanf("%s",op);
            if(op[0]=='R')
            {
                if(top)updata(q[top--],1,1,n,1);
            }
            else scanf("%d",&i);
            if(op[0]=='D')
            {
                q[++top]=i;
                updata(i,0,1,n,1);
            }
            if(op[0]=='Q')printf("%d\n",query(i,1,n,1));
        }
    }
    return 0;
}