提供两种思路
一种线段树区间更新
另一种用map维护连续的区间,也是题解的思路
第二种很难写(我太渣,看了别人的代码,发现自己写的太烦了)

#include<iostream>
#include<map>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<set>
#include<vector>
#include<queue>
#include<stack>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const int N = 6e5+5;
#define MS(x,y) memset(x,y,sizeof(x))
#define MP(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1 int main() {
int n, q;
while(~scanf("%d %d", &n, &q)) {
map<int, int> mp;
mp[-1] = -1;
mp[1] = n;
mp[n+1] = INF; int ans = n;
while(q --) {
int l, r, k; scanf("%d %d %d", &l, &r, &k);
int tt = r;
if(k == 2 && r != n) tt = r+1; auto it = mp.upper_bound(l);
auto it2 = mp.upper_bound(tt);
it --; it2 --; int head1 = it -> first; int len1 = it -> second;
int head2 = it2 -> first; int len2 = it2 -> second; while(1){
int flag = 0;
if(it == it2) flag = 1;
ans -= it->second;
auto tmp = it;
// if(it->first == n+1) while(1);
it ++;
// printf("erase: %d\n", tmp->first);
mp.erase(tmp);
if(flag) break;
}
//for(auto i = mp.begin(); i != mp.end(); ++i) printf("%d:%d ", i->first, i->second); printf("\n");
// printf("%d %d %d %d %d %d\n", head1, len1, head2, len2, l, r); if(k == 1) {
if(head1 + len1 - 1 >= l && l!=head1) {
mp[head1] = l - head1;
ans += l - head1;
} else if(l != head1){
mp[head1] = len1;
ans += len1;
} if(head2 + len2 - 1 > r) {
mp[r+1] = head2 + len2 - 1 - r;
ans += head2 + len2 - 1 - r;
}
} else {
int L = l; int R = max(r, head2 + len2 -1);
if(head1 + len1 < l) {
mp[head1] = len1;
ans += len1;
}else L = head1; mp[L] = R-L+1;
ans += R-L+1;
} // for(auto i = mp.begin(); i != mp.end(); ++i) printf("%d:%d ", i->first, i->second); printf("\n"); printf("%d\n", ans);
}
}
return 0;
}
#include<iostream>
#include<map>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<set>
#include<vector>
#include<queue>
#include<stack>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const int N = 6e5+5;
#define MS(x,y) memset(x,y,sizeof(x))
#define MP(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1 int Q[N][3];
int has[N * 2]; int cnt; int sum[N << 2];
int lazy[N << 2]; void build(int l, int r, int rt) {
lazy[rt] = 0;
sum[rt] = has[r] - has[l-1];
if(l == r) {
return;
}
int m = (l + r) >> 1;
build(lson); build(rson);
}
void update(int ty, int L, int R, int l, int r, int rt) {
//printf("%d %d\n", l, r);
if(L <= has[l-1]+1 && has[r] <= R) {
lazy[rt] = ty == 1? -1 : 1;
sum[rt] = ty == 1? 0 : has[r] - has[l-1];
return;
} int m = (l + r) >> 1; if(lazy[rt] == 1) {
lazy[rt<<1] = 1; sum[rt<<1] = has[m] - has[l-1];
lazy[rt<<1|1] = 1; sum[rt<<1|1] = has[r] - has[m];
lazy[rt] = 0;
} else if(lazy[rt] == -1){
lazy[rt<<1] = -1; sum[rt<<1] = 0;
lazy[rt<<1|1] = -1; sum[rt<<1|1] = 0;
lazy[rt] = 0;
} if(L <= has[m-1]+1) update(ty, L, R, lson);
if(R > has[m]) update(ty, L, R, rson);
sum[rt] = sum[rt<<1] + sum[rt<<1|1];
}
void debug(int l, int r, int rt) {
printf("%d %d %d\n", l, r, sum[rt]);
if(l == r) return;
int m = (l + r) >> 1;
debug(lson);
debug(rson);
}
int main() {
int n;
while(~scanf("%d", &n)) {
cnt = 0; int q; scanf("%d", &q);
for(int i = 0; i < q; ++i) {
scanf("%d %d %d", &Q[i][0], &Q[i][1], &Q[i][2]);
has[cnt ++] = Q[i][0] - 1;
has[cnt ++] = Q[i][1];
}
has[cnt ++] = 0;
has[cnt ++] = n; sort(has, has+cnt);
cnt = unique(has, has + cnt) - has;
// for(int i = 0; i < cnt; ++i) printf("%d ", has[i]); printf("\n"); build(1, cnt-1, 1);
//` debug(1, cnt-1, 1);
for(int i = 0; i < q; ++i) {
update(Q[i][2], Q[i][0], Q[i][1], 1, cnt-1, 1);
printf("%d\n", sum[1]);
// debug(1, cnt-1, 1);
} }
return 0;
}

Educational Codeforces Round 36 (Rated for Div. 2) E. Physical Education Lessons的更多相关文章

  1. Educational Codeforces Round 36 (Rated for Div. 2)

    A. Garden time limit per test 1 second memory limit per test 256 megabytes input standard input outp ...

  2. Educational Codeforces Round 36 (Rated for Div. 2) G. Coprime Arrays

    求a_i 在 [1,k]范围内,gcd(a_1,a_2...,a_n) = 1的a的数组个数. F(x)表示gcd(a_1,a_2,...,a_n) = i的a的个数 f(x)表示gcd(a_1,a_ ...

  3. Educational Codeforces Round 71 (Rated for Div. 2)-E. XOR Guessing-交互题

    Educational Codeforces Round 71 (Rated for Div. 2)-E. XOR Guessing-交互题 [Problem Description] ​ 总共两次询 ...

  4. Educational Codeforces Round 60 (Rated for Div. 2) - C. Magic Ship

    Problem   Educational Codeforces Round 60 (Rated for Div. 2) - C. Magic Ship Time Limit: 2000 mSec P ...

  5. Educational Codeforces Round 60 (Rated for Div. 2) - D. Magic Gems(动态规划+矩阵快速幂)

    Problem   Educational Codeforces Round 60 (Rated for Div. 2) - D. Magic Gems Time Limit: 3000 mSec P ...

  6. Educational Codeforces Round 43 (Rated for Div. 2)

    Educational Codeforces Round 43 (Rated for Div. 2) https://codeforces.com/contest/976 A #include< ...

  7. Educational Codeforces Round 35 (Rated for Div. 2)

    Educational Codeforces Round 35 (Rated for Div. 2) https://codeforces.com/contest/911 A 模拟 #include& ...

  8. Codeforces Educational Codeforces Round 44 (Rated for Div. 2) F. Isomorphic Strings

    Codeforces Educational Codeforces Round 44 (Rated for Div. 2) F. Isomorphic Strings 题目连接: http://cod ...

  9. Codeforces Educational Codeforces Round 44 (Rated for Div. 2) E. Pencils and Boxes

    Codeforces Educational Codeforces Round 44 (Rated for Div. 2) E. Pencils and Boxes 题目连接: http://code ...

随机推荐

  1. Java:对象的强、软、弱和虚引用[转]

    原文链接:http://zhangjunhd.blog.51cto.com/113473/53092/ 原创作品,允许转载,转载时请务必以超链接形式标明文章 原始出处 .作者信息和本声明.否则将追究法 ...

  2. CodeChef Chef and Churu [分块]

    题意: 单点修改$a$ 询问$a$的区间和$f$的区间和 原来普通计算机是这道题改编的吧... 对$f$分块,预处理$c[i][j]$为块i中$a_j$出现几次,$O(NH(N))$,只要每个块差分加 ...

  3. vue2.0路由进阶

    一.路由的模式 第一种用history方式实现,HTML5使用window.history.pushState()实现路由的切换而不刷新页面. 第二种使用hash值的方式来实现. vue2.0两种都可 ...

  4. AMD && CMD

    前言 JavaScript初衷:实现简单的页面交互逻辑,寥寥数语即可: 随着web2.0时代的到来,Ajax技术得到广泛应用,jQuery等前端库层出不穷,前端代码日益膨胀 问题: 这时候JavaSc ...

  5. JDBC常见面试题

    以下我是归纳的JDBC知识点图: 图上的知识点都可以在我其他的文章内找到相应内容. JDBC常见面试题 JDBC操作数据库的步骤 ? JDBC操作数据库的步骤 ? 注册数据库驱动. 建立数据库连接. ...

  6. python爬虫(3)——SSL证书与Handler处理器

    一.SSL证书问题 上一篇文章,我们创建了一个小爬虫,下载了上海链家房产的几个网页.实际上我们在使用urllib联网的过程中,会遇到证书访问受限的问题. 处理HTTPS请求SSL证书验证,如果SSL证 ...

  7. href与src 区别

    src 是可替换的文本支撑,将指向的内容引入文档当前标签所在的位置, 当浏览器解析到该标签时,将暂停其它资源的下载处理, 请求该标签的src ,下载指向的外部资源并应用到当前文档, 所以js 脚本一般 ...

  8. VMware Workstation All Key

    官方下载:https://www.vmware.com/products/workstation-pro/workstation-pro-evaluation.html 懒人打包:链接:https:/ ...

  9. python爬取快手视频 多线程下载

    就是为了兴趣才搞的这个,ok 废话不多说 直接开始. 环境: python 2.7 + win10 工具:fiddler postman 安卓模拟器 首先,打开fiddler,fiddler作为htt ...

  10. mysql 学习心得1

    1由于不靠这玩意吃饭 估计不准备精读 顺便中文版也不用担心翻译问题 科科 大致翻了下=,= mysql的感觉怎么就是背命令.... 2DDL语句 定义数据 创建删除修改 create drop alt ...