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

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

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 ;
 }