CodeForces 145E Lucky Queries (线段树)

题意：给定一个47序列，然后有两种操作，

1.switch l, r 把区间内的4变成7，7变成4

2.count 计算整个区间的最长的非降序序列长度。

析：一个很裸的线段树，就是维护几个值，一个是只有4的长度，一个只有7的，一个47都有的，每个都维护正着反着，然后就很简单了。

在更新时，4和7直接相加就好了，47都有的取个最大值，左边4右边7，左边4，右边47，左边47右边7。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e6 + 10;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}

int length4[maxn<<2][2], length7[maxn<<2][2], length[maxn<<2][2];
bool rev[maxn<<2];
char s[maxn];

void push_up(int rt){
  int l = rt<<1, r = rt<<1|1;
  for(int i = 0; i < 2; ++i){
    length4[rt][i] = length4[l][i] + length4[r][i];
    length7[rt][i] = length7[l][i] + length7[r][i];
    length[rt][i] = max(length4[l][i]+length[r][i], max(length[l][i]+length7[r][i], length4[l][i]+length7[r][i]));
  }
}

void rever(int rt){
  swap(length4[rt][0], length4[rt][1]);
  swap(length7[rt][0], length7[rt][1]);
  swap(length[rt][0], length[rt][1]);
  rev[rt] = !rev[rt];
}

void push_down(int rt){
  if(!rev[rt])  return ;
  rever(rt<<1);
  rever(rt<<1|1);
  rev[rt] = 0;
}

void build(int l, int r, int rt){
  if(l == r){
    s[l] == '4' ? ++length4[rt][0] : ++length7[rt][0];
    length4[rt][1] = length7[rt][0];
    length7[rt][1] = length4[rt][0];
    return ;
  }
  int m = l + r >> 1;
  build(lson);
  build(rson);
  push_up(rt);
}

void update(int L, int R, int l, int r, int rt){
  if(L <= l && r <= R){
    rever(rt);
    return ;
  }
  push_down(rt);
  int m = l + r >> 1;
  if(L <= m)  update(L, R, lson);
  if(R > m)   update(L, R, rson);
  push_up(rt);
}

int query(int rt){
  int ans = max(length4[rt][0], length7[rt][0]);
  return max(ans, length[rt][0]);
}

int main(){
  scanf("%d %d", &n, &m);
  scanf("%s", s+1);
  build(1, n, 1);
  while(m--){
    scanf("%s", s);
    if(s[0] == 's'){
      int l, r;
      scanf("%d %d", &l, &r);
      update(l, r, 1, n, 1);
    }
    else  printf("%d\n", query(1));
  }
  return 0;
}