BZOJ1230 [Usaco2008 Nov]lites 开关灯

区间not，求区间1的个数。。。线段树裸题

然而窝并不会线段树

我们可以对序列分块，每个块记录0/1的个数和tag表示又没有区间not过就好了

 /**************************************************************
     Problem: 1230
     User: rausen
     Language: C++
     Result: Accepted
     Time:796 ms
     Memory:1608 kb
 ****************************************************************/

 #include <cstdio>
 #include <cmath>
 #include <algorithm>

 using namespace std;
 const int N = 1e5 + ;
 const int SZ = ;

 int n, m;
 int a[N];
 int sz, b[N], st[SZ], cnt[SZ][], tag[SZ], cnt_block;

 inline int read() {
     static int x;
     static char ch;
     x = , ch = getchar();
     while (ch < '' || '' < ch)
         ch = getchar();
     while ('' <= ch && ch <= '') {
         x = x *  + ch - '';
         ch = getchar();
     }
     return x;
 }

 inline void pre() {
     int i;
     sz = sqrt(n);
     if (!sz) sz = ;
     for (i = ; i <= n; ++i) {
         if (i % sz ==  || sz == ) st[++cnt_block] = i;
         b[i] = cnt_block, ++cnt[cnt_block][];
     }
     st[cnt_block + ] = n + ;
 }

 inline void push_tag(int B) {
     static int i;
     if (!tag[B]) return;
     for (i = st[B]; i < st[B + ]; ++i)
         a[i] = !a[i];
     swap(cnt[B][], cnt[B][]);
     tag[B] = ;
 }

 inline void change(int x, int y) {
     static int i, t1, t2;
     t1 = b[x], t2 = b[y];
     for (i = t1 + ; i < t2; ++i) tag[i] = !tag[i];
     if (t1 == t2)
         for (push_tag(t1), i = x; i <= y; ++i)
             --cnt[t1][a[i]], a[i] = !a[i], ++cnt[t1][a[i]];
     else {
         for (push_tag(t1), i = x; i < st[t1 + ]; ++i)
             --cnt[t1][a[i]], a[i] = !a[i], ++cnt[t1][a[i]];
         for (push_tag(t2), i = st[t2]; i <= y; ++i)
             --cnt[t2][a[i]], a[i] = !a[i], ++cnt[t2][a[i]];
     }
 }

 inline int query(int x, int y) {
     static int i, t1, t2, res;
     t1 = b[x], t2 = b[y], res = ;
     for (i = t1 + ; i < t2; ++i)
         res += cnt[i][!tag[i]];
     if (t1 == t2)
         for (push_tag(t1), i = x; i <= y; ++i) res += a[i];
     else {
         for (push_tag(t1), i = x; i < st[t1 + ]; ++i) res += a[i];
         for (push_tag(t2), i = st[t2]; i <= y; ++i) res += a[i];
     }
     return res;
 }

 int main() {
     int i, oper, x, y;
     n = read(), m = read();
     pre();
     for (i = ; i <= m; ++i) {
         oper = read(), x = read(), y = read();
         if (oper == ) change(x, y);
         else printf("%d\n", query(x, y));
     }
     return ;
 }