HDU 3911 Black And White （线段树区间更新）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3911

题面：

Black And White

Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4358    Accepted Submission(s): 1271

Problem Description

There are a bunch of stones on the beach; Stone color is white or black. Little Sheep has a magic brush, she can change the color of a continuous stone, black to white, white to black. Little Sheep like black very much, so she want to know the longest period of consecutive black stones in a range [i, j].

 

Input

  There are multiple cases, the first line of each case is an integer n(1<= n <= 10^5), followed by n integer 1 or 0(1 indicates black stone and 0 indicates white stone), then is an integer M(1<=M<=10^5) followed by M operations formatted as x i j(x = 0 or 1) , x=1 means change the color of stones in range[i,j], and x=0 means ask the longest period of consecutive black stones in range[i,j]

 

Output

When x=0 output a number means the longest length of black stones in range [i,j].

 

Sample Input

41 0 1 050 1 41 2 30 1 41 3 30 4 4

 

Sample Output

120

 

Source

2011 Multi-University Training Contest 8 - Host by HUST

题目大意：

    初始给定一01序列，两种操作，1 a b，取反a到b间的元素，0 a b，问a b区间内最长的连续1的长度。

解题：

    第一颗区间更新的线段树，来得有点晚啊，借鉴、理解也算是写出来了。题意看似简单，合并操作比较复杂。需要维护5个值。maxw,maxb,le,ri,lazy。

    maxw/maxb，该区间的最长连续0/1长度，取反时直接交换两值。

    le/ri，该区间的左右连续块长度，用于合并。

    lazy该区间是否需要继续向下更新。

代码：

#include <iostream>#include <cmath>#include <cstdio>#define siz 100010//左右子树编号#define ls i<<1#define rs (i<<1)|1using namespace std;//左右边界，该区间最多连续黑块数，白块数，因为反转区间需要//区间左边界连续块数，右边解连续块数，正表黑，负表白，用于区间合并//lazy标记如果为1，代表更新到该区间，其下区间尚未更新struct SEGTree{int l,r,maxb,maxw,le,ri,lazy;}STree[siz<<2];int a[siz];//取大int max(int x,int y){return x>y?x:y;}//取小int min(int a,int b){  return a<b?a:b;}//向上更新void push_up(int i){   int tmp;   //取左右区间最大黑块数   int maxb=max(STree[ls].maxb,STree[rs].maxb);   //如果可以合并的话，将合并的值一同比较   if(STree[ls].ri>0&&STree[rs].le>0)   {   tmp=STree[ls].ri+STree[rs].le;   if(tmp>maxb)   maxb=tmp;   }   //更新   STree[i].maxb=maxb;   //白块更新同上   int maxw=max(STree[ls].maxw,STree[rs].maxw);   if(STree[ls].ri<0&&STree[rs].le<0)   {   tmp=STree[ls].ri+STree[rs].le;   tmp=-tmp;   if(tmp>maxw)   maxw=tmp;   }   STree[i].maxw=maxw;   //区间左端最长连续块数le，如果左子区单色，且能与右子区相连，更新le   if(abs(STree[ls].le)==(STree[ls].r-STree[ls].l+1)&&(STree[ls].ri*STree[rs].le>0))   STree[i].le=STree[ls].le+STree[rs].le;   //否则直接取左子区le   else STree[i].le=STree[ls].le;   //更新ri，同上   if(abs(STree[rs].ri)==(STree[rs].r-STree[rs].l+1)&&(STree[ls].ri*STree[rs].le>0))   STree[i].ri=STree[ls].ri+STree[rs].ri;   else STree[i].ri=STree[rs].ri;}//向下更新void push_down(int i){//该区间的懒标记清零STree[i].lazy=0;//更新左右子区间swap(STree[ls].maxw,STree[ls].maxb);STree[ls].le=-STree[ls].le;STree[ls].ri=-STree[ls].ri;STree[ls].lazy=!STree[ls].lazy;swap(STree[rs].maxw,STree[rs].maxb);STree[rs].le=-STree[rs].le;STree[rs].ri=-STree[rs].ri;STree[rs].lazy=!STree[rs].lazy;}//建树void build(int i,int l,int r){STree[i].l=l;STree[i].r=r;STree[i].lazy=0;//叶子节点if(l==r){//如果是黑块   if(a[l])   {   STree[i].maxb=1;   STree[i].maxw=0;   STree[i].le=1;   STree[i].ri=1;   }   //白块   else   {   STree[i].maxw=1;   STree[i].maxb=0;   STree[i].le=-1;   STree[i].ri=-1;   }   return;}//向下递归int mid=(l+r)>>1;build(i<<1,l,mid);build((i<<1)|1,mid+1,r);//向上更新push_up(i);}void update(int i,int l,int r){//如果刚好区间吻合if(STree[i].l==l&&STree[i].r==r){//黑白块交换swap(STree[i].maxw,STree[i].maxb);STree[i].le=-STree[i].le;STree[i].ri=-STree[i].ri;//lazy取反STree[i].lazy=!STree[i].lazy;return;}//如果没有刚好吻合，要访问的区间为当前区间子区间，//且该区间还有懒标记未下放，则向下更新，并向下访问if(STree[i].lazy)push_down(i);int mid=(STree[i].l+STree[i].r)>>1;if(r<=mid)update(ls,l,r);else if(l>mid)update(rs,l,r);else{update(ls,l,mid);update(rs,mid+1,r);}//更新之后，向上更新push_up(i);}int query(int i,int l,int r){//如果刚好吻合，则返回区间最大连续黑块if(l==STree[i].l&&r==STree[i].r)  return STree[i].maxb;int mid=(STree[i].l+STree[i].r)>>1;//不能完全符合，且该区间尚有懒标记，下放懒标记if(STree[i].lazy)push_down(i);    //向下查询if(r<=mid)  return query(ls,l,r);else if(l>mid)  return query(rs,l,r);else{int res,tmp;//取左右最值，合并的最值res=max(query(ls,l,mid),query(rs,mid+1,r));//此处不仅仅合并左右区间最值，还需限制左右边界不能超出查询区间以mid为中心的左右边界        tmp=min(mid-l+1,STree[ls].ri)+min(r-mid,STree[rs].le);if(tmp>res)  res=tmp;return res;}}int main(){    int n,q,oper,fm,to;while(~scanf("%d",&n)){  //读入  for(int i=1;i<=n;i++)    scanf("%d",&a[i]);  //建树      build(1,1,n);  scanf("%d",&q);  for(int i=1;i<=q;i++)  {  scanf("%d",&oper);          scanf("%d%d",&fm,&to);  if(oper)update(1,fm,to);  else          {              int ans=query(1,fm,to);              printf("%d\n",ans);          }  }}return 0;}