hdu 1540 Tunnel Warfare(线段树区间统计)

Tunnel Warfare

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3293    Accepted Submission(s): 1272

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

 

题意：

是一条线上的点，D x是破坏这个点，Q x是表示查询以x所在的最长的连续的点的个数，R是恢复上一次破坏的点。

思路：

这是一道典型的线段树区间统计类题目。用线段树维护两个值ml,mr。ml表示结点对应区间左连续点(即连通村子)的个数。mr区间右连续点的个数。对于D和R操作都很简单。单点修改递归更新就行了。对于R操作用一个栈记录摧毁村子的标号就行。有点难的是询问操作。就是询问一个点所在连续区间元素的个数。

一个点对应于线段树区间只有3种情况。

1.在左连续区间内。

2.在右连续区间内。

3.夹在这两个区间内。

分情况处理就好。

只是要注意。这题有点坑。discuss里面说

1）多case
（2）某个村庄可以被毁坏多次（必须全部入栈），但只需要一次就能将其恢复（下面的这组数据，，）
（3）D 3  D 2  D 1  D 1  D 2
    R  恢复2
    R  恢复1
    R  恢复3

其实根本不用。这么做反而wa了。不用考虑那么多D就直接入栈。R就直接出栈就行了。

详细见代码：

#include <iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
const int maxn=50010;
int ml[maxn<<2],mr[maxn<<2],sta[maxn];//sta为栈
int flag,tail;
void btree(int L,int R,int k)
{
    int ls,rs,mid;
    ml[k]=mr[k]=R-L+1;
    if(L==R)
        return;
    ls=k<<1;
    rs=ls|1;
    mid=(L+R)>>1;
    btree(L,mid,ls);
    btree(mid+1,R,rs);
}
void update(int L,int R,int k,int p,int op)//更新很简单
{
    int ls,rs,mid;
    if(L==R)
    {
        if(op)
            ml[k]=mr[k]=0;
        else
            ml[k]=mr[k]=1;
        return;
    }
    ls=k<<1;
    rs=ls|1;
    mid=(L+R)>>1;
    if(p>mid)
        update(mid+1,R,rs,p,op);
    else
        update(L,mid,ls,p,op);
    ml[k]=ml[ls];
    mr[k]=mr[rs];
    if(ml[ls]==mid-L+1)
        ml[k]+=ml[rs];
    if(mr[rs]==R-mid)
        mr[k]+=mr[ls];
}
int qu(int L,int R,int k,int p)
{
    int ls,rs,mid,t;
    if(mr[k]>=R-p+1)
    {
        flag=1;//加以标记。返回上层时加上扩展值
        return mr[k];
    }
    if(ml[k]>=p-L+1)
    {
        flag=1;
        return ml[k];
    }
    if(L==R)
        return 0;
    ls=k<<1;
    rs=ls|1;
    mid=(L+R)>>1;
    if(p>mid)
    {
        t=qu(mid+1,R,rs,p);
        if(flag)
        {
            flag=0;
            t+=mr[ls];
        }
    }
    else
    {
        t=qu(L,mid,ls,p);
        if(flag)
        {
            flag=0;
            t+=ml[rs];
        }
    }
    return t;
}
int main()
{
    int n,m,p,i;
    char com[10];

    while(~scanf("%d%d",&n,&m))
    {
        btree(1,n,1);
        flag=tail=0;
        for(i=0;i<m;i++)
        {
            scanf("%s",com);
            switch(com[0])
            {
            case 'D':
                scanf("%d",&p);
                update(1,n,1,p,1);
                sta[tail++]=p;
                break;
            case 'Q':
                scanf("%d",&p);
                flag=0;//注意flag初始化！
                printf("%d\n",qu(1,n,1,p));
                break;
            case 'R':
                update(1,n,1,sta[--tail],0);
            }
        }
    }
    return 0;
}
/*
3 100
D 3
D 1
D 2
R
Q 1
Q 2
Q 3
R
Q 1
Q 2
Q 3
*/