题解 P2574 【XOR的艺术】

主要思路：线段树

线段树大法好

我觉得这道题就是把区间修改，区间查询的普通线段树改了改懒标记就完了

不会线段树？不着急啊，我们有入门宝典——

具体线段树入门：

入门1：单点修改，区间查询

入门2：懒标记及区间修改

Blog发完就跑

记得，这里的xor如果xor两次就相当于没操作，所以我们在维护懒标记时可以再偷点懒，每次懒标记取反。

我们区间取反时，实际上是原来是0，现在变为1，原来是1，现在变为0，他们的数目是互补的，所以区间取反后1的数目就是之前0的数目

代码：

#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
using namespace std;
#define go(i, j, n, k) for (int i = j; i <= n; i += k)
#define fo(i, j, n, k) for (int i = j; i >= n; i -= k)
#define rep(i, x) for (int i = h[x]; i; i = e[i].nxt)
#define mn 222222
#define inf 2147483647
#define ll long long
#define ld long double
#define fi first
#define se second
#define root 1, n, 1
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
#define bson l, r, rt
//#define LOCAL
#define mod 
#define Debug(...) fprintf(stderr, __VA_ARGS__)
inline int read(){
    int f = 1, x = 0;char ch = getchar();
    while (ch > '9' || ch < '0'){if (ch == '-')f = -f;ch = getchar();}
    while (ch >= '0' && ch <= '9'){x = x * 10 + ch - '0';ch = getchar();}
    return x * f;
}
//This is AC head above...
int z[mn << 2], col[mn << 2], a[mn];
inline void update(int rt){
    z[rt] = z[rt << 1] + z[rt << 1 | 1];
}
inline void color(int l,int r,int rt){
    z[rt] = (r - l + 1) - z[rt];
    col[rt] ^= 1;
}
inline void push_col(int l,int r,int rt){
    if(col[rt]){
        int m = (l + r) >> 1;
        color(lson);
        color(rson);
        col[rt] = 0;
    }
}
inline void build(int l,int r,int rt){
    if(l==r){
        z[rt] = a[l];
        return;
    }
    int m = (l + r) >> 1;
    build(lson);
    build(rson);
    update(rt);
}
inline void modify(int l,int r,int rt,int nowl,int nowr){
    if(nowl<=l && r<=nowr){
        col[rt] ^= 1;
        z[rt] = (r - l + 1) - z[rt];
        return;
    }
    int m = (l + r) >> 1;
    push_col(bson);
    if(nowl<=m)
        modify(lson, nowl, nowr);
    if(m<nowr)
        modify(rson, nowl, nowr);
    update(rt);
}
inline int query(int l,int r,int rt,int nowl,int nowr){
    if(nowl<=l && r<=nowr){
        return z[rt];
    }
    int m = (l + r) >> 1;
    push_col(bson);
    if(nowl<=m){
        if(m<nowr)
            return query(lson, nowl, nowr) + query(rson, nowl, nowr);
        else
            return query(lson, nowl, nowr);
    }else
        return query(rson, nowl, nowr);
}
int n, m;
inline void debug(){
    go(i, 1, n, 1) printf("%d ", a[i]);
    puts("");
}
int main(){
    n = read(), m = read();
    string c;
    cin >> c;
    go(i, 0, n - 1, 1)
        a[i + 1] = c[i] - '0';
    //debug();
    build(root);
    go(i,1,m,1){
        int s = read(), x = read(), y = read();
        if(!s)
            modify(root, x, y);
        else
            cout << query(root, x, y) << "
";
    }
#ifdef LOCAL
    Debug("
My Time: %.3lfms
", (double)clock() / CLOCKS_PER_SEC);
#endif
    return 0;
}

第十三次发题解，希望可以帮助初学者入门线段树