hdu 3911 Black and White 线段树维护01序列

Black And White

Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4996 Accepted Submission(s): 1509

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
4
1 0 1 0
5
0 1 4
1 2 3
0 1 4
1 3 3
0 4 4

Sample Output
1
2
0

题意：给你一些01序列，再给你一些询问。当询问为1的时候，把给定区间内所有的0变为1，1边为0，当询问为0的时候，输出所求区间中最长的连续的1的个数。

用线段树。
维护的东西：左右区间最长的1序列和左右区间最长的0序列。左区间最右端有几个连续的1/0，右区间有几个连续的1/0，以及lazy标记，以及最长的0/1序列的答案。
每次比较左区间最大值和右区间的最大值，以及左区间最右端和右区间最左端的和，最大的作为大区间的答案。
每次lazy标记下传就交换01串的答案就行了。
我果然还是只会线段树的傻逼题。。。

#include<cstdio>#include<algorithm>#include<cstring>using namespace std;const int N = 1000010;int a[N];inline int Max(int a,int b){    return a>b?a:b;}struct node{    int mmax1,mmax0;    int flag;    int lm1,rm1;//分别表示左区间最长的1的串和右区间的最长的1的串     int lm0,rm0;    int l,r;}tree[N];void pushup(int root){    int llen=tree[root<<1].r-tree[root<<1].l+1;    int rlen=tree[root<<1|1].r-tree[root<<1|1].l+1;    tree[root].lm1=tree[root<<1].lm1;    if(tree[root<<1].lm1==llen) tree[root].lm1+=tree[root<<1|1].lm1;    tree[root].rm1=tree[root<<1|1].rm1;    if(tree[root<<1|1].rm1==rlen) tree[root].rm1+=tree[root<<1].rm1;    int c=Max(tree[root<<1].mmax1,tree[root<<1|1].mmax1);    int d=tree[root<<1|1].lm1+tree[root<<1].rm1;    tree[root].mmax1=Max(c,d);    tree[root].lm0=tree[root<<1].lm0;    if(tree[root<<1].lm0==llen) tree[root].lm0+=tree[root<<1|1].lm0;    tree[root].rm0=tree[root<<1|1].rm0;    if(tree[root<<1|1].rm0==rlen) tree[root].rm0+=tree[root<<1].rm0;    c=Max(tree[root<<1].mmax0,tree[root<<1|1].mmax0);    d=tree[root<<1|1].lm0+tree[root<<1].rm0;    tree[root].mmax0=Max(c,d);}void pushdown(int root){    int flag = tree[root].flag;    if(flag){        tree[root<<1].flag ^= 1;        tree[root<<1|1].flag ^= 1;        swap(tree[root<<1].mmax1,tree[root<<1].mmax0);        swap(tree[root<<1].lm1,tree[root<<1].lm0);        swap(tree[root<<1].rm1,tree[root<<1].rm0);        swap(tree[root<<1|1].mmax1,tree[root<<1|1].mmax0);        swap(tree[root<<1|1].lm1,tree[root<<1|1].lm0);        swap(tree[root<<1|1].rm1,tree[root<<1|1].rm0);        tree[root].flag=0;      }}void build(int root,int l,int r){    tree[root].l=l,tree[root].r=r;    if(l==r){        if(a[l]==1){           tree[root].mmax1=tree[root].lm1=tree[root].rm1=a[l]&1;           tree[root].mmax0=tree[root].lm0=tree[root].rm0=a[l]^1;        }        else{           tree[root].mmax1=tree[root].lm1=tree[root].rm1=0;//a[l]&1;           tree[root].mmax0=tree[root].lm0=tree[root].rm0=1;//a[l]^1;                   }        tree[root].flag=0;        return ;    }    int mid=(l+r)>>1;    build(root<<1,l,mid);    build(root<<1|1,mid+1,r);    tree[root].flag=0;    pushup(root);}void modify(int root,int pos,int val){    int l=tree[root].l,r=tree[root].r;    if(pos==l&&val==r){        tree[root].flag ^= 1;        swap(tree[root].mmax1,tree[root].mmax0);        swap(tree[root].lm1,tree[root].lm0);        swap(tree[root].rm1,tree[root].rm0);        return ;    }    int mid=l+r>>1;    pushdown(root);    if(val<=mid) modify(root<<1,pos,val);    else if(pos>mid) modify(root<<1|1,pos,val);    else{        modify(root<<1,pos,mid);        modify(root<<1|1,mid+1,val);    }    pushup(root);}int query(int root,int pos,int val){    int l=tree[root].l,r=tree[root].r;    if(l==pos&&r==val) return tree[root].mmax1;    pushdown(root);    int mid=(l+r)>>1;    if(val<=mid) return query(root<<1,pos,val);    if(pos>mid) return query(root<<1|1,pos,val);    int ll=query(root<<1,pos,mid);    int rl=query(root<<1|1,mid+1,val);    int a=tree[root<<1].rm1;    if(a>tree[root<<1].r-pos+1) a=tree[root<<1].r-pos+1;    int b=tree[root<<1|1].lm1;    if(b>val-tree[root<<1|1].l+1) b=val-tree[root<<1|1].l+1;    int ans = Max(ll,Max(rl,a+b));    return ans; }int main(){    int n,m,x,l,r;    while(~scanf("%d",&n)){      for(register int i=1;i<=n;i++)          scanf("%d",&a[i]);      build(1,1,n);      scanf("%d",&m);      for(register int i=1;i<=m;i++){          scanf("%d%d%d",&x,&l,&r);          if(x==0) printf("%d\n",query(1,l,r));          else if(x==1) modify(1,l,r);      }    }    return 0;}