【HDU

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
可以提前先把这个题写了

题意：有n个数和m此操作，操作分两种——
1，0 x y 表示把区间[x, y]里面的1改为0，0改为1；
2，1 x y 表示查询区间[x, y]最多的连续的1的个数。
代码

#include<bits/stdc++.h>using namespace std;typedef pair<int,int>pii;#define first fi#define second se#define  LL long long#define fread() freopen("in.txt","r",stdin)#define fwrite() freopen("out.txt","w",stdout)#define CLOSE() ios_base::sync_with_stdio(false)const int MAXN = 1e5+11;const int MAXM = 1e6;const int mod = 1e9+7;const int inf = 0x3f3f3f3f;struct Tree{    int l,r,len;// 同时维护区间的连续1的个数，和0的个数。    int l1,r1,s1;    int l0,r0,s0;    int lazy;}tree[MAXN<<2];void Up(int o){    tree[o].l1=tree[o<<1].l1;  //维护1    tree[o].r1=tree[o<<1|1].r1;    tree[o].s1=max(tree[o<<1].s1,tree[o<<1|1].s1);    if(tree[o<<1].l1==tree[o<<1].len)         tree[o].l1+=tree[o<<1|1].l1;    if(tree[o<<1|1].r1==tree[o<<1|1].len)         tree[o].r1+=tree[o<<1].r1;    tree[o].s1=max(tree[o].s1,max(tree[o].l1,tree[o].r1));    tree[o].s1=max(tree[o].s1,tree[o<<1].r1+tree[o<<1|1].l1);    tree[o].l0=tree[o<<1].l0;//维护0    tree[o].r0=tree[o<<1|1].r0;    tree[o].s0=max(tree[o<<1].s0,tree[o<<1|1].s0);    if(tree[o<<1].l0==tree[o<<1].len)         tree[o].l0+=tree[o<<1|1].l0;    if(tree[o<<1|1].r0==tree[o<<1|1].len)         tree[o].r0+=tree[o<<1].r0;    tree[o].s0=max(tree[o].s0,max(tree[o].l0,tree[o].r0));    tree[o].s0=max(tree[o].s0,tree[o<<1].r0+tree[o<<1|1].l0);}void Change(int o){// 如果操作0时候，就是相当于把0和1的数据交换就可以了。    swap(tree[o].l1,tree[o].l0);    swap(tree[o].r1,tree[o].r0);    swap(tree[o].s1,tree[o].s0);    tree[o].lazy^=1; //  lazy是延迟标记，如果有2次lazy到这个区间，就相当于没有变化，所以用的^}void Down(int o){    if(tree[o].lazy){        Change(o<<1);        Change(o<<1|1);        tree[o].lazy=0;    }}void Build(int o,int le,int ri){    tree[o].l=le,tree[o].r=ri,tree[o].len=ri-le+1;    tree[o].lazy=0;    if(le==ri){        int val;scanf("%d",&val);        tree[o].l1=tree[o].r1=tree[o].s1=val?1:0;        tree[o].l0=tree[o].r0=tree[o].s0=val?0:1;        return ;    }     int mid=(le+ri)>>1;    Build(o<<1,le,mid);    Build(o<<1|1,mid+1,ri);    Up(o);}void Update(int o,int le,int ri){    if(le<=tree[o].l&&tree[o].r<=ri) {        Change(o);        return ;    }    Down(o);    int mid=(tree[o].l+tree[o].r)>>1;    if(ri<=mid) Update(o<<1,le,ri);    else if(le>mid) Update(o<<1|1,le,ri);    else {        Update(o<<1,le,mid);        Update(o<<1|1,mid+1,ri);    }     Up(o);}int Query(int o,int le,int ri){    if(tree[o].l>=le&&tree[o].r<=ri) {        return tree[o].s1;    }    Down(o);    int mid=(tree[o].l+tree[o].r)>>1;    if(ri<=mid) return Query(o<<1,le,ri);    else if(le>mid) return Query(o<<1|1,le,ri);    else {         int ans=Query(o<<1,le,mid);         ans=max(ans,Query(o<<1|1,mid+1,ri));         return ans=max(ans,min(mid-le+1,tree[o<<1].r1)+min(ri-mid,tree[o<<1|1].l1));    }}int main(){    CLOSE();//  fread();//  fwrite();    int n;    while(scanf("%d",&n)!=EOF){        Build(1,1,n);        int m;scanf("%d",&m);        while(m--){            int op,a,b;            scanf("%d%d%d",&op,&a,&b);            if(op)                Update(1,a,b);            else                 printf("%d\n",Query(1,a,b));        }    }    return 0;}