﻿ hdu 1540 Tunnel Warfare （线段树区间合并）-C 语言_软件编程-脚本宝典

# Tunnel Warfare

Time Limit: 4000/2000 MS (/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 6065 Accepted Submission(s): 2344

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
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```#include
#include
#include

using namespace std;

struct node
{
int l,r;
int ls,rs,ms;//分别表示左边最大连续，右边最大连续，以及整个区间内的最大连续长度
} s[50050*3];

int n,m;
int op[50010];

void InitTree(int l,int r,int k)
{
s[k].l=l;
s[k].r=r;
s[k].ls=s[k].rs=s[k].ms=r-l+1;//最开始的时候全都是连着的。所以长度为r-l+1
if (l==r)
return ;
int mid=(l+r)/2;
InitTree(l,mid,k*2);
InitTree(mid+1,r,k*2+1);
}

void UpdataTree(int x,int flag,int k)//x表示修复或者破坏的数字，flag用来标记是破坏还是修复
{
if (s[k].l==s[k].r)
{
if (flag==1)
s[k].ls=s[k].rs=s[k].ms=1;//修复
else
s[k].ls=s[k].rs=s[k].ms=0;//破坏
return ;
}
int mid=(s[k].l+s[k].r)/2;
if (x<=mid)
UpdataTree(x,flag,2*k);
else
UpdataTree(x,flag,2*k+1);
if(s[2*k].ls == s[2*k].r-s[2*k].l+1)//左区间的左连续=左子树的长度，就说名左区间的数全部连续，（左子树区间满了），整个区间的左区间就应该加上有区间的左部分。
s[k].ls =s[2*k].ls+s[2*k+1].ls;
else
s[k].ls=s[2*k].ls;
if(s[2*k+1].rs==s[2*k+1].r-s[2*k+1].l+1)//同理
s[k].rs=s[2*k+1].rs+s[2*k].rs;
else
s[k].rs=s[2*k+1].rs;
s[k].ms=max(max(s[2*k].ms,s[2*k+1].ms),s[2*k].rs+s[2*k+1].ls);//整个区间内的最大连续应为：左子树最大区间，右子树最大区间，左右子树合并的中间区间，三者中取最大
}

int SearchTree(int x,int k)
{
if(s[k].l==s[k].r||s[k].ms==0||s[k].ms==s[k].r-s[k].l+1)//到了叶子节点或者该访问区间为空或者已满都不必要往下走了
return s[k].ms;
int mid=(s[k].l+s[k].r)/2;
if (x<=mid)
{
if (x>=s[2*k].r-s[2*k].rs+1)//判断当前这个数是否在左区间的右连续中，其中s[2*k].r-s[2*k].rs+1代表左子树右边连续区间的左边界值，即有连续区间的起点
return s[2*k].rs+s[2*k+1].ls;//也可以SearchTree(x,2*k)+SearchTree(mid+1,2*k+1);
else
return SearchTree(x,2*k);
}
else
{
if (x<=s[2*k+1].l+s[2*k+1].ls-1)//判断当前这个数是否在左区间的右连续中，其中s[2*k].r-s[2*k].rs+1代表左子树右边连续区间的左边界值，即有连续区间的起点
return s[2*k].rs+s[2*k+1].ls;//这种方法SearchTree(x,2*k+1)+SearchTree(mid,2*k);也是可以的，但是比较浪费时间
else
return SearchTree(x,2*k+1);
}
}

int main()
{
int x;
char ch[2];
while (~scanf ("%d%d",&n,&m))
{
int top=0;
InitTree(1,n,1);
while (m--)
{
scanf("%s",ch);
if (ch[0]=='D')
{
scanf("%d",&x);
op[top++]=x;
UpdataTree(x,0,1);
}
else if (ch[0]=='Q')
{
scanf("%d",&x);
printf ("%d\n",SearchTree(x,1));
}
else
{
if (x>0)
{
x=op[--top];
UpdataTree(x,1,1);
}
}
}
}
return 0;
}
```

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