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

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

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int maxn = 100010;
 4 struct node {
 5     int lsum[2],rsum[2],mx[2];
 6     bool lazy;
 7 } tree[maxn<<2];
 8 void pushup(int v,int k) {
 9     for(int i = 0; i < 2; ++i) {
10         tree[v].lsum[i] = tree[v<<1].lsum[i];
11         tree[v].rsum[i] = tree[v<<1|1].rsum[i];
12         if(tree[v].lsum[i] == k - (k>>1))
13             tree[v].lsum[i] += tree[v<<1|1].lsum[i];
14         if(tree[v].rsum[i] == (k>>1))
15             tree[v].rsum[i] += tree[v<<1].rsum[i];
16         tree[v].mx[i] = max(tree[v<<1].mx[i],tree[v<<1|1].mx[i]);
17         tree[v].mx[i] = max(tree[v].mx[i],tree[v<<1].rsum[i]+tree[v<<1|1].lsum[i]);
18     }
19 }
20 
21 void change(int v) {
22     swap(tree[v].lsum[0],tree[v].lsum[1]);
23     swap(tree[v].rsum[0],tree[v].rsum[1]);
24     swap(tree[v].mx[0],tree[v].mx[1]);
25     tree[v].lazy = !tree[v].lazy;
26 }
27 void pushdown(int v) {
28     if(tree[v].lazy) {
29         change(v<<1);
30         change(v<<1|1);
31         tree[v].lazy = false;
32     }
33 }
34 void build(int L,int R,int v) {
35     tree[v].lazy = false;
36     if(L == R) {
37         int tmp;
38         scanf("%d",&tmp);
39         tree[v].mx[0] = tree[v].lsum[0] = tree[v].rsum[0] = !tmp;
40         tree[v].mx[1] = tree[v].lsum[1] = tree[v].rsum[1] = tmp;
41         return;
42     }
43     int mid = (L + R)>>1;
44     build(L,mid,v<<1);
45     build(mid+1,R,v<<1|1);
46     pushup(v,R - L + 1);
47 }
48 void update(int L,int R,int lt,int rt,int v) {
49     if(lt <= L && rt >= R) {
50         change(v);
51         return;
52     }
53     pushdown(v);
54     int mid = (L + R)>>1;
55     if(lt <= mid) update(L,mid,lt,rt,v<<1);
56     if(rt > mid) update(mid+1,R,lt,rt,v<<1|1);
57     pushup(v,R - L + 1);
58 }
59 int query(int L,int R,int lt,int rt,int v) {
60     if(lt <= L && rt >= R) return tree[v].mx[1];
61     pushdown(v);
62     int mid = (L + R)>>1,ret = 0;
63     if(lt <= mid) ret = max(ret,query(L,mid,lt,rt,v<<1));
64     if(rt > mid) ret = max(ret,query(mid+1,R,lt,rt,v<<1|1));
65     if(lt <= mid && rt > mid)
66         ret = max(ret,min(mid - lt + 1,tree[v<<1].rsum[1]) + min(rt - mid,tree[v<<1|1].lsum[1]));
67     pushup(v,R - L + 1);
68     return ret;
69 }
70 int main() {
71     int n,m,op,a,b;
72     while(~scanf("%d",&n)){
73         build(1,n,1);
74         scanf("%d",&m);
75         while(m--){
76             scanf("%d %d %d",&op,&a,&b);
77             if(op) update(1,n,a,b,1);
78             else printf("%d
",query(1,n,a,b,1));
79         }
80     }
81     return 0;
82 }

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