poj 2892 Tunnel Warfare（线段树…

题意：抗日时期，八路军擅用地道把村落连起来，现指挥官要知道一些信息：对地道的操作如下

D x 表示销毁x这个地道

Q x 表示查询有多少个地道与x相连

R 修复最后被摧毁的地道

思路：线段路，记录每个结点左边 中间，右边 连续的地道数量。

//3412K
266MS

#include <stdio.h>

#define M 50050

#define L(x) (x<<1)

#define R(x) ((x<<1)+1)

struct data

{

int
l,r,state; //state
-1表示空，1表示全被destroy

int
lma,mma,rma; //左中右连续的数量

}node[3*M];

int D[M];

void BuildTree(int left ,int right,int u)

{

node[u].l =
left;

node[u].r =
right;

if (left ==
right)

return ;

int mid =
(left + right)>>1;

BuildTree(left,mid,L(u));

BuildTree(mid+1,right,R(u));

}

void getdown(int u,int op)

{

node[u].state = 0;

if (op ==
-1)

{

node[L(u)].state = -1;

node[L(u)].lma = node[L(u)].mma = node[L(u)].rma = node[L(u)].r -
node[L(u)].l + 1;

node[R(u)].state = -1;

node[R(u)].lma = node[R(u)].mma = node[R(u)].rma = node[R(u)].r -
node[R(u)].l + 1;

}

else if (op
== 1)

{

node[L(u)].state = 1;

node[L(u)].lma = node[L(u)].mma = node[L(u)].rma = 0;

node[R(u)].state = 1;

node[R(u)].lma = node[R(u)].mma = node[R(u)].rma = 0;

}

}

void Union (int u)

{

if
(node[L(u)].state == -1)

node[u].lma = (node[L(u)].r - node[L(u)].l + 1) +
node[R(u)].lma;

else

node[u].lma = node[L(u)].lma;

if
(node[R(u)].state == -1)

node[u].rma = (node[R(u)].r - node[R(u)].l + 1) +
node[L(u)].rma;

else

node[u].rma = node[R(u)].rma;

node[u].mma
= node[L(u)].rma + node[R(u)].lma; //

}

void destroy (int num,int u) //销毁操作

{

if
(node[u].state == 1)

return ;

if
(node[u].l == num&&node[u].r ==
num) //找到点修改它的值

{

node[u].state = 1;

node[u].lma = node[u].rma = node[u].mma = 0;

return ;

}

if
(node[u].state == -1) //延迟覆盖

getdown(u,-1);

if (num
<= node[L(u)].r)

destroy(num,L(u));

if (num
>= node[R(u)].l)

destroy(num,R(u));

Union
(u);
//根据左右结点，修改父结点

if
(node[L(u)].state == node[R(u)].state)

node[u].state = node[L(u)].state;

}

int query (int num,int u)
//查询操作

{

if
(node[u].state == -1)

return node[u].r - node[u].l + 1;

if
(node[u].l == num&&node[u].r ==
num) //如果它的状态不为-1 则一定是被销毁了，返回0

return 0;

if (num
>= node[u].l && num
<= node[u].l+node[u].lma-1)//如果在左连续的段内

return node[u].lma;

if (num
<= node[u].r &&num
>= node[u].r - node[u].rma + 1)//在右

return node[u].rma;

if (num
>=
node[L(u)].r-node[L(u)].rma+1&&num
<= node[R(u)].l + node[R(u)].lma - 1)//在中间

return node[u].mma;

if (num
<= node[L(u)].r)

query (num,L(u));

else if (num
>= node[R(u)].l)

query (num,R(u));

}

void rebuild (int num ,int u)

{

// if (node[u].state ==
-1)

//
return ;

if
(node[u].l == num&&node[u].r ==
num)

{

node[u].state = -1;

node[u].lma = node[u].rma = node[u].mma = 1;

return ;

}

if
(node[u].state == 1)

getdown(u,1);

if (num
<= node[L(u)].r)

rebuild(num,L(u));

if (num
>= node[R(u)].l)

rebuild(num,R(u));

Union
(u);

if
(node[L(u)].state == node[R(u)].state)

node[u].state = node[L(u)].state;

}

int main ()

{

int
n,m,num;

char
op;

while
(~scanf ("%d%d",&n,&m))

{

BuildTree(1,n,1);

node[1].state = -1;

node[1].lma = node[1].rma = node[1].mma = node[1].r-node[1].l +
1;

int i = 0;

while (m --)

{

getchar ();

scanf ("%c",&op);

if (op == 'D')

{

scanf ("%d",&num);

D[i++] = num;

destroy (num,1);

}

else if (op == 'Q')

{

scanf ("%d",&num);

int len = query (num,1);

printf ("%d\n",len);

}

else

{

num = D[--i];

rebuild (num,1);

}

}

}

return
0;

}