hdu1540 Tunnel Warfare

Tunnel Warfare

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7703    Accepted Submission(s):
2981

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

题目大意：

有n个村庄，每个操作：

D x 表示摧毁x

Q x 询问与x相连接的有多少个

R 恢复之前摧毁的一个城市

这道题是线段树的区间合并。

首先用0表示已经摧毁的村庄，1表示未摧毁或已经修复的村庄

记录的信息有两个：

1.l表示该区间从左数的最长1串

2.r表示该区间从右数的最长1串

pushup时在长度没有覆盖整个左区间或右区间时直接做

tr[x].l=tr[ls].l;
tr[x].r=tr[rs].r;

如果完整覆盖了左或右区间再做特殊处理

if(tr[ls].l==mid-l+1) tr[x].l+=tr[rs].l;
if(tr[rs].r==r-mid) tr[x].r+=tr[ls].r;

所以对于摧毁和恢复直接进行单点修改，将tr[x].l=tr[x].r=z(z为0或1)

对于询问，可以分别查询a向左的最长1串和a向右最长1串-1

特别的如果a已经被摧毁，则ans=-1

此时要做特判。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define ls x<<1
#define rs x<<1|1
const int N=;
struct X
{
    int l,r;
}tr[N<<];
int st[N];
void bu(int l,int r,int x)
{
    if(l==r) tr[x].l=tr[x].r=;
    else
    {
        int mid=l+(r-l)/;
        bu(l,mid,ls);
        bu(mid+,r,rs);
        tr[x].l=tr[x].r=r-l+;
    }
}
void chan(int l,int r,int x,int w,int z)
{
    if(l==r) tr[x].l=tr[x].r=z;
    else
    {
        int mid=l+(r-l)/;
        if(w<=mid) chan(l,mid,ls,w,z);
        else chan(mid+,r,rs,w,z);
        tr[x].l=tr[ls].l;tr[x].r=tr[rs].r;
　　　　 if(tr[x].l==mid-l+) tr[x].l+=tr[rs].l;
        if(tr[x].r==r-mid) tr[x].r+=tr[ls].r;
    }
}
int ask1(int l,int r,int x,int w)
{
    if(w==r)  return tr[x].r;
    else
    {
        int mid=l+(r-l)/,re;
        if(w>mid)
        {
            re=ask1(mid+,r,rs,w);
            if(w-mid==re) re+=tr[ls].r;
        }
        else re=ask1(l,mid,ls,w);
        return re;
    }
}
int ask2(int l,int r,int x,int w)
{
    if(w==l)  return tr[x].l;
    else
    {
        int mid=l+(r-l)/,re;
        if(w<=mid)
        {
            re=ask2(l,mid,ls,w);
            if(mid-w+==re) re+=tr[rs].l;
        }
        else re=ask2(mid+,r,rs,w);
        return re;
    }
}
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        int s=;
        memset(st,,sizeof(st));
        memset(tr,,sizeof(tr));
        bu(,n,);
        while(m--)
        {
            char c;
            scanf("\n%c",&c);
            if(c=='D')
            {
                scanf("%d",&st[++s]);
                chan(,n,,st[s],);
            }
            else if(c=='R')    chan(,n,,st[s--],);
            else
            {
                int a,ans;
                scanf("%d",&a);
                ans=ask1(,n,,a)+ask2(,n,,a)-;
                if(ans>) printf("%d\n",ans);
                else printf("0\n");
            }
        }
    }
    return ;
}