HDU-1540 Tunnel Warfare 线段树 区间合并

http://acm.hdu.edu.cn/showproblem.php?pid=1540

题意： 
给出直线上一系列的村庄，如果相邻村庄都没有被破坏， 
则两村庄是连接的，题目给出一系列的破坏操作，对指定号码的村庄进行破坏， 
还有一系列的询问操作，询问与指定号码的村庄直接相连或间接相连的村庄有几个， 
还有一个修复操作，是对最后破坏的村庄进行修复。 


//线段树
定义三个变量：lmax[rt] 表示从左孩子开始的连续区间个数 
              rmax[rt] 到右孩子截至的连续区间个数 
              mmax[rt] 整个区间连续的区间个数 
              col[rt] 表示区间的状态，0表示未必破坏，1表示已被破坏，-1初始状态 

因为为连续区间关键步骤是pushdown 里的赋值操作和pushup的更新操作 

#include "stdio.h"#include "string.h"const int maxn = 50005;int n,q,x,val;int stack[maxn],top;bool vis[maxn];struct node                     {    int col;              //表示区间的状态，0表示未必破坏，1表示已被破坏，-1初始状态     int rmax,lmax,mmax;   //到右孩子截至的连续区间个数  表示从左孩子开始的连续区间个数 整个区间连续的区间个数 }tree[maxn*4];int Max( int a,int b ){    return a>=b?a:b;}void pushup( int ld,int rd,int t ){    int mid = ( ld+rd )>>1;    tree[t].lmax = tree[t<<1].lmax;    tree[t].rmax = tree[t<<1|1].rmax;    tree[t].mmax = Max( tree[t<<1].mmax,tree[t<<1|1].mmax );        if( tree[t<<1].lmax == mid - ld +1 )    {        tree[t].lmax += tree[t<<1|1].lmax;    }    if( tree[t<<1|1].rmax == rd - mid )    {        tree[t].rmax += tree[t<<1].rmax;    }    tree[t].mmax = Max( tree[t].mmax,tree[t<<1].rmax+tree[t<<1|1].lmax);}void pushdown( int ld,int rd,int t ){    if( tree[t].col!=-1 )    {        int mid = ( ld+rd )>>1;        tree[t<<1].col = tree[t<<1|1].col =tree[t].col;        if( tree[t].col==1 )        {            tree[t<<1].lmax = tree[t<<1].rmax = tree[t<<1].mmax = 0;            tree[t<<1|1].lmax = tree[t<<1|1].rmax = tree[t<<1|1].mmax = 0;        }        else        {            tree[t<<1].lmax = tree[t<<1].rmax = tree[t<<1].mmax = mid-ld+1;            tree[t<<1|1].lmax = tree[t<<1|1].rmax = tree[t<<1|1].mmax = rd-mid;        }        tree[t].col = -1;    }}void buildtree( int ld,int rd,int t ){    tree[t].col = -1;    tree[t].rmax = tree[t].lmax = tree[t].mmax = rd-ld+1;     if( ld == rd )        return;    int mid = ( ld+rd )>>1;    buildtree( ld,mid,t<<1 );    buildtree( mid+1,rd,t<<1|1 );}void updata( int ld,int rd,int t ){    if( ld == rd )    {        tree[t].col = val;        tree[t].lmax = tree[t].rmax = tree[t].mmax = ( val?0:1 );        return;    }    pushdown( ld,rd,t );    int mid = ( ld+rd )>>1;    if( x<=mid )        updata( ld,mid,t<<1 );    else        updata( mid+1,rd,t<<1|1 );    pushup( ld,rd,t );}int query( int ld,int rd,int t ){    if( ld == rd || tree[t].mmax==( rd-ld+1 ) || !tree[t].mmax )    {        return tree[t].mmax;    }    int mid = ( ld+rd )>>1;    if( x<=mid )    {        if( x >= mid - tree[t<<1].rmax+1 )            return tree[t<<1].rmax + tree[t<<1|1].lmax;        else            return query( ld,mid,t<<1 );    }    else    {        if( x <= mid + tree[t<<1|1].lmax+1 )            return tree[t<<1].rmax + tree[t<<1|1].lmax;        else            return query( mid+1,rd,t<<1|1 );    }}int main(){    int i;    char ch[3];    while( scanf("%d%d",&n,&q)==2 )    {        top = 0;        memset(vis,0,sizeof(vis));        getchar();        buildtree( 1,n,1 );        for( i=1;i<=q;i++ )        {            scanf("%s",ch);            if( ch[0]=='D' )            {                scanf("%d",&x);                stack[++top] = x;                                val = 1;//1为摧毁                if( !vis[x] )                    updata(1,n,1);                vis[x] = 1;            }            else if( ch[0]=='Q' )            {                scanf("%d",&x);                printf("%d\n",vis[x]?0:query(1,n,1) );            }            else            {                if( top )                {                    x = stack[top--];                                        val = 0;//0为修复                                        if( vis[x] )                        updata( 1,n,1 );                    vis[x] = 0;                }            }        }    }    return 0;}