I - Tunnel Warfare HDU 1540（線段樹+最大的連續區間）

I - Tunnel Warfare

Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 1540

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

2

4

http://www.cnblogs.com/scau20110726/archive/2013/05/07/3065418.html

這篇文章的分析很好，但是查詢沒有說，

#include<bits/stdc++.h>
//#include<iostream>
//#include<cmath>
//#include<cstring>
//#include<string>
//#include<cstdlib>
//#include<cstdio>
//#include<map>
//#include<algorithm>
using namespace std;
template<class T>inline T read(T&x)
{
    char c;
    while((c=getchar())<=32)if(c==EOF)return -1;
    bool ok=false;
    if(c=='-')ok=true,c=getchar();
    for(x=0; c>32; c=getchar())
        x=x*10+c-'0';
    if(ok)x=-x;
    return 1;
}
template<class T> inline T read_(T&x,T&y)
{
    return read(x)!=-1&&read(y)!=-1;
}
template<class T> inline T read__(T&x,T&y,T&z)
{
    return read(x)!=-1&&read(y)!=-1&&read(z)!=-1;
}
template<class T> inline void write(T x)
{
    if(x<0)putchar('-'),x=-x;
    if(x<10)putchar(x+'0');
    else write(x/10),putchar(x%10+'0');
}
template<class T>inline void writeln(T x)
{
    write(x);
    putchar('\n');
}
//-------ZCC IO template------
const int maxn=50010;
const double inf=999999999;
#define lson (rt<<1),L,M
#define rson (rt<<1|1),M+1,R
#define M ((L+R)>>1)
#define For(i,t,n) for(int i=(t);i<(n);i++)
typedef long long  LL;
typedef double DB;
#define bug printf("---\n");

struct node
{
    int tlen,llen,rlen;//tlen表示目前區間的最大連續長度，llen表示目前區間左端點的連續長度，rlen表示目前區間右端點的連續長度
}p[maxn<<2];
int stak[maxn],top;//本題棧很簡單，直接用數組模拟即可

void build(int rt,int L,int R)
{
    p[rt].llen=p[rt].rlen=p[rt].tlen=R-L+1;//建立樹，開始時都是連續的
    if(L==R)return ;
    build(lson);
    build(rson);
}

void update(int rt,int L,int R,int i,int v)
{
    if(L==R)
    {
        p[rt].llen=p[rt].rlen=p[rt].tlen=v;//到達葉子結點直接更新
        return ;
    }

    if(i<=M)update(lson,i,v);
    else update(rson,i,v);

    p[rt].llen=p[rt<<1].llen;//維護父親的rlen，llen，tlen，畫個圖就能了解
    p[rt].rlen=p[rt<<1|1].rlen;
    p[rt].tlen=max(max(p[rt<<1].tlen,p[rt<<1|1].tlen),p[rt<<1].rlen+p[rt<<1|1].llen);

    if(p[rt<<1].tlen==M-L+1)//關鍵的地方，如果左兒子區間全部是連續的，那麼他父親的左端點的連續長度就要加上該父親右兒子左端點的連續長度
        p[rt].llen+=p[rt<<1|1].llen;
    if(p[rt<<1|1].tlen==R-M)//右兒子全部連續，父親的右端點的連續長度，就要加上左兒子的右端點的連續長度
        p[rt].rlen+=p[rt<<1].rlen;//如果覺得繞，畫個圖，很清晰
}

int query(int rt,int L,int R,int pos)
{
    if(L==R||p[rt].tlen==0||p[rt].tlen==R-L+1)
    {
        return p[rt].tlen;
    }
    if(pos<=M)
    {
        if(p[rt<<1].rlen>=M-pos+1)//如果查詢到POS在的左兒子區間是全部連續的，那麼就要加上右兒子的左端點的連續，下面同理
            return query(lson,pos)+query(rson,M+1);
        else
            return query(lson,pos);
    }
    else
    {
        if(p[rt<<1|1].llen>=pos-M)
            return query(rson,pos)+query(lson,M);
        else
            return query(rson,pos);
    }
}

int main()
{
    //freopen("in.txt","r",stdin);
    int n,m,i,j,k,t;
    while(read_(n,m))
    {
        build(1,1,n);
        char op[2];
        top=0;
        while(m--)
        {
            scanf("%s",op);
            if(op[0]=='D')
            {
                read(k);
                stak[top++]=k;
                update(1,1,n,k,0);
            }
            else if(op[0]=='Q')
            {
                read(k);
                writeln(query(1,1,n,k));
            }
            else
            {
                k=stak[--top];
                update(1,1,n,k,1);
            }
        }
    }
    return 0;
}