ZOJ 2301 Color the Ball (离散化+线段树)

题意：有从 1 开始递增依次编号的很多球，开始他们都是黑色的，现在依次给出 n 个操作（ai，bi，ci），每个操作都是把编号 ai 到 bi 区间内

的-所有球涂成 ci 表示的颜色（黑 or 白），然后经过 n 次给定的操作后，求最长的连续白色区间的左端点和右端点。

析：由于数比较大，可以先进行离散化，然后是区间更新，最后还循环暴力一次，单点求值。

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 = 2000 + 5;
const int mod = 1000000007;
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;
}

struct Node{
  LL l, r;
  int c;
};
Node a[maxn];
vector<LL> v;
map<LL, int> mp;

int sum[maxn<<4];
int setv[maxn<<4];

void push_down(int rt){
  if(setv[rt] == -1)  return ;
  setv[rt<<1] = setv[rt];
  setv[rt<<1|1] = setv[rt];
  sum[rt] = setv[rt];
  sum[rt<<1] = setv[rt];
  sum[rt<<1|1] = setv[rt];
  setv[rt] = -1;
}

void update(int L, int R, int val, int l, int r, int rt){
  if(L <= l && r <= R){
    setv[rt] = val;
    sum[rt] = val;
    return ;
  }
  push_down(rt);
  int m = l + r >> 1;
  if(L <= m)  update(L, R, val, lson);
  if(R > m)  update(L, R, val, rson);
}

int query(int M, int l, int r, int rt){
  if(l == r)  return sum[rt];
  push_down(rt);
  int m = l + r >> 1;
  if(M <= m)  return query(M, lson);
  return query(M, rson);
}

int main(){
  while(scanf("%d", &n) == 1){
    v.clear();
    mp.clear();
    char s[5];
    for(int i = 0; i < n; ++i){
      scanf("%d %d %s", &a[i].l, &a[i].r, s);
      a[i].c = s[0] == 'b' ? 0 : 1;
      v.push_back(a[i].l);
      v.push_back(a[i].r);
      v.push_back(a[i].l-1);
      v.push_back(a[i].r-1);
      v.push_back(a[i].l+1LL);
      v.push_back(a[i].r+1LL);
    }
    sort(v.begin(), v.end());
    v.erase(unique(v.begin(), v.end()), v.end());
    for(int i = 0; i < v.size(); ++i)
      mp[v[i]] = i + 1;
    memset(sum, 0, sizeof sum);
    memset(setv, -1, sizeof setv);
    int len = v.size();
    for(int i = 0; i < n; ++i){
      if(a[i].l > a[i].r)  continue;
      update(mp[a[i].l], mp[a[i].r], a[i].c, 1, len, 1);
    }
    LL l = -1, r = -1, ans = 0;
    LL ll = -1, rr = -1;
    for(int i = 0; i < len; ++i){
      if(query(i+1, 1, len, 1)){
        if(ll == -1)  ll = v[i];
        rr = v[i];
        if(ans < rr - ll + 1){
          ans = rr - ll + 1;
          l = ll, r = rr;
        }
        
      }
      else  ll = rr = -1;
    }

    if(ans == 0) cout << "Oh, my god
";
    else   cout << l << " " << r << endl;
  }
  return 0;
}