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
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\n",query(1,n,a,b,1));
79 }
80 }
81 return 0;
82 }