HDU 3911 Black And White（线段树区间合并）

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

 

线段树区间合并：对于每个区间[l,r]所对应的rs,我们给出6个量:

lmax_1:左起的1的最大前缀,lmax_0:左起的0的最大前缀；

rmax_1:右起的1的最大后缀rmax_0:右起的0的最大后缀;

max_1：区间的1的最大连续值，max_0：区间的0的最大连续值

我们的任务就是不断地更新求值。

#include<cstdio>#include<cstring>#include<algorithm>#include<vector>#include<string>#include<iostream>#include<queue>#include<cmath>#include<map>#include<stack>#include<bitset>using namespace std;#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )#define CLEAR( a , x ) memset ( a , x , sizeof a )typedef long long LL;typedef pair<int,int>pil;const int mod = 1000000007;const int maxn=1e5+10;int lmax_1[maxn<<2],rmax_1[maxn<<2];int lmax_0[maxn<<2],rmax_0[maxn<<2];int max_1[maxn<<2],max_0[maxn<<2];int col[maxn<<2];int n,m,os;void pushup(int rs,int m){    lmax_1[rs]=lmax_1[rs<<1];//处理左边一类前缀    lmax_0[rs]=lmax_0[rs<<1];    if(lmax_1[rs<<1]==m-(m>>1))  lmax_1[rs]+=lmax_1[rs<<1|1];    if(lmax_0[rs<<1]==m-(m>>1))  lmax_0[rs]+=lmax_0[rs<<1|1];    rmax_1[rs]=rmax_1[rs<<1|1];//处理右边一类后缀    rmax_0[rs]=rmax_0[rs<<1|1];    if(rmax_1[rs<<1|1]==(m>>1))  rmax_1[rs]+=rmax_1[rs<<1];    if(rmax_0[rs<<1|1]==(m>>1))  rmax_0[rs]+=rmax_0[rs<<1];    max_1[rs]=max(max_1[rs<<1],max_1[rs<<1|1]);//处理整个区间的值    max_1[rs]=max(max_1[rs],rmax_1[rs<<1]+lmax_1[rs<<1|1]);    max_0[rs]=max(max_0[rs<<1],max_0[rs<<1|1]);    max_0[rs]=max(max_0[rs],rmax_0[rs<<1]+lmax_0[rs<<1|1]);}void work(int rs){    swap(lmax_1[rs],lmax_0[rs]);    swap(rmax_1[rs],rmax_0[rs]);    swap(max_1[rs],max_0[rs]);}void push_down(int rs){    if(col[rs])    {        col[rs<<1]^=1;col[rs<<1|1]^=1;        col[rs]=0;work(rs<<1);work(rs<<1|1);    }}void build(int rs,int l,int r){    col[rs]=0;    if(l==r)    {        scanf("%d",&os);        if(os==1)        {            lmax_1[rs]=rmax_1[rs]=max_1[rs]=1;            lmax_0[rs]=rmax_0[rs]=max_0[rs]=0;        }        else        {            lmax_1[rs]=rmax_1[rs]=max_1[rs]=0;            lmax_0[rs]=rmax_0[rs]=max_0[rs]=1;        }        return ;    }    int mid=(l+r)>>1;    build(rs<<1,l,mid);    build(rs<<1|1,mid+1,r);    pushup(rs,r-l+1);}void update(int x,int y,int l,int r,int rs){    if(l>=x&&r<=y)    {        col[rs]^=1;        work(rs);//交换值        return ;    }    push_down(rs);    int mid=(l+r)>>1;    if(x<=mid)  update(x,y,l,mid,rs<<1);    if(y>mid)  update(x,y,mid+1,r,rs<<1|1);    pushup(rs,r-l+1);}int query(int x,int y,int l,int r,int rs){    if(l>=x&&r<=y)        return max_1[rs];    push_down(rs);    int mid=(l+r)>>1;    if(x>mid)  return query(x,y,mid+1,r,rs<<1|1);    if(y<=mid)  return query(x,y,l,mid,rs<<1);    int t1=query(x,y,l,mid,rs<<1);    int t2=query(x,y,mid+1,r,rs<<1|1);    int r1=min(mid-x+1,rmax_1[rs<<1]);//以mid为分割点    int r2=min(y-mid,lmax_1[rs<<1|1]);    int res=max(max(t1,t2),r1+r2);    return res;}int main(){    int op,x,y;    while(~scanf("%d",&n))    {       build(1,1,n);       scanf("%d",&m);       while(m--)       {           scanf("%d%d%d",&op,&x,&y);           if(op==1)               update(x,y,1,n,1);           else               printf("%d\n",query(x,y,1,n,1));       }    }    return 0;}