HDU 3911 Black And White

Black And White

Time Limit: 3000ms

Memory Limit: 32768KB

This problem will be judged on HDU. Original ID: 3911
64-bit integer IO format: %I64d Java class name: Main

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

Sample Output

Source

2011 Multi-University Training Contest 8 - Host by HUST

解题：线段树啊。。。幸好只有一种操作，翻转啊。。。

就是给你两种操作，1 a b表示将a b区里面的01翻转，0 a b表示输出a b间连续的最长的1有多少个

 #include <bits/stdc++.h>

 using namespace std;

 const int maxn = ;

 struct node {

     int lsum[],rsum[],mx[];

     bool lazy;

 } tree[maxn<<];

 void pushup(int v,int k) {

     for(int i = ; i < ; ++i) {

         tree[v].lsum[i] = tree[v<<].lsum[i];

         tree[v].rsum[i] = tree[v<<|].rsum[i];

         if(tree[v].lsum[i] == k - (k>>))

             tree[v].lsum[i] += tree[v<<|].lsum[i];

         if(tree[v].rsum[i] == (k>>))

             tree[v].rsum[i] += tree[v<<].rsum[i];

         tree[v].mx[i] = max(tree[v<<].mx[i],tree[v<<|].mx[i]);

         tree[v].mx[i] = max(tree[v].mx[i],tree[v<<].rsum[i]+tree[v<<|].lsum[i]);

     }

 }

 void change(int v) {

     swap(tree[v].lsum[],tree[v].lsum[]);

     swap(tree[v].rsum[],tree[v].rsum[]);

     swap(tree[v].mx[],tree[v].mx[]);

     tree[v].lazy = !tree[v].lazy;

 }

 void pushdown(int v) {

     if(tree[v].lazy) {

         change(v<<);

         change(v<<|);

         tree[v].lazy = false;

     }

 }

 void build(int L,int R,int v) {

     tree[v].lazy = false;

     if(L == R) {

         int tmp;

         scanf("%d",&tmp);

         tree[v].mx[] = tree[v].lsum[] = tree[v].rsum[] = !tmp;

         tree[v].mx[] = tree[v].lsum[] = tree[v].rsum[] = tmp;

         return;

     }

     int mid = (L + R)>>;

     build(L,mid,v<<);

     build(mid+,R,v<<|);

     pushup(v,R - L + );

 }

 void update(int L,int R,int lt,int rt,int v) {

     if(lt <= L && rt >= R) {

         change(v);

         return;

     }

     pushdown(v);

     int mid = (L + R)>>;

     if(lt <= mid) update(L,mid,lt,rt,v<<);

     if(rt > mid) update(mid+,R,lt,rt,v<<|);

     pushup(v,R - L + );

 }

 int query(int L,int R,int lt,int rt,int v) {

     if(lt <= L && rt >= R) return tree[v].mx[];

     pushdown(v);

     int mid = (L + R)>>,ret = ;

     if(lt <= mid) ret = max(ret,query(L,mid,lt,rt,v<<));

     if(rt > mid) ret = max(ret,query(mid+,R,lt,rt,v<<|));

     if(lt <= mid && rt > mid)

         ret = max(ret,min(mid - lt + ,tree[v<<].rsum[]) + min(rt - mid,tree[v<<|].lsum[]));

     pushup(v,R - L + );

     return ret;

 }

 int main() {

     int n,m,op,a,b;

     while(~scanf("%d",&n)){

         build(,n,);

         scanf("%d",&m);

         while(m--){

             scanf("%d %d %d",&op,&a,&b);

             if(op) update(,n,a,b,);

             else printf("%d\n",query(,n,a,b,));

         }

     }

     return ;

 }

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