CF368 D - Persistent Bookcase
re了20多发 还是我在测试数据上操作最后了10多发才发现的
其实只需要多加一句就好了
真的愚蠢啊,要不都能进前100了
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN = 100005;
const int INF = 0x3f3f3f3f; int sh[1005][1005];
int flag[1005];
int sum[1005];
int all;
int vis[MAXN];
vector<int> mp[MAXN];
int n,m,q; struct Node{
int ty; int x,y; int ans;
}as[MAXN]; int cc = 0;
const int Ed = 26;
void dfs(int po,int fa) { int x1,x2; int y1,y2,y3,y4; // sh[x1][x2], all, sum[x1], flag[x1]
for(int i = 0; i < (int)mp[po].size(); ++i) {
int tt = mp[po][i];
int t1 = as[tt].ty; int t2 = as[tt].x; int t3 = as[tt].y; if(t1 == 4) t2 = 0; // 这地方不写会越界 谨记啊
x1 = t2; x2 = t3; y1 = sh[x1][x2]; y2 = all; y3 = sum[x1]; y4 = flag[x1]; if(t1 == 1) {
if( (flag[t2] ^ sh[t2][t3]) == 0) {
all ++; sh[t2][t3] ^= 1; sum[t2] ++;
}
}else if(t1 == 2) {
if( (flag[t2] ^ sh[t2][t3]) == 1){
all--; sh[t2][t3] ^= 1; sum[t2] --;
}
}else if(t1 == 3) {
flag[t2] ^= 1;
all += m-sum[t2]*2;
sum[t2] = m-sum[t2];
} as[tt].ans = all;
dfs(tt,po); sh[x1][x2] = y1; all = y2; sum[x1] = y3; flag[x1] = y4;
}
}
int main(){
while(~scanf("%d %d %d",&n,&m,&q)){
memset(flag,0,sizeof(flag));
memset(sh,0,sizeof(sh));
memset(sum,0,sizeof(sum));
all = 0;
cc = 0; for(int i = 0; i <= q; ++i) mp[i].clear(); for(int i = 1; i <= q; ++i) {
as[i].y = 0;
scanf("%d",&as[i].ty);
if(as[i].ty < 3) scanf("%d %d",&as[i].x,&as[i].y);
else scanf("%d",&as[i].x); if(as[i].ty == 4) mp[as[i].x].push_back(i);
else mp[i-1].push_back(i);
} dfs(0,-1);
for(int i = 1; i <= q; ++i) printf("%d\n",as[i].ans);
}
return 0;
}
CF368 D - Persistent Bookcase的更多相关文章
- CodeForces #368 div2 D Persistent Bookcase DFS
题目链接:D Persistent Bookcase 题意:有一个n*m的书架,开始是空的,现在有k种操作: 1 x y 这个位置如果没书,放书. 2 x y 这个位置如果有书,拿走. 3 x 反转这 ...
- 【Codeforces-707D】Persistent Bookcase DFS + 线段树
D. Persistent Bookcase Recently in school Alina has learned what are the persistent data structures: ...
- Codeforces Round #368 (Div. 2) D. Persistent Bookcase
Persistent Bookcase Problem Description: Recently in school Alina has learned what are the persisten ...
- Persistent Bookcase
Persistent Bookcase time limit per test 2 seconds memory limit per test 512 megabytes input standard ...
- Codeforces Round #368 (Div. 2) D. Persistent Bookcase 离线 暴力
D. Persistent Bookcase 题目连接: http://www.codeforces.com/contest/707/problem/D Description Recently in ...
- codeforces 707D D. Persistent Bookcase(dfs)
题目链接: D. Persistent Bookcase time limit per test 2 seconds memory limit per test 512 megabytes input ...
- CF707D Persistent Bookcase
CF707D Persistent Bookcase 洛谷评测传送门 题目描述 Recently in school Alina has learned what are the persistent ...
- Persistent Bookcase CodeForces - 707D (dfs 离线处理有根树模型的问题&&Bitset)
Persistent Bookcase CodeForces - 707D time limit per test 2 seconds memory limit per test 512 megaby ...
- D. Persistent Bookcase(Codeforces Round #368 (Div. 2))
D. Persistent Bookcase time limit per test 2 seconds memory limit per test 512 megabytes input stand ...
随机推荐
- MongoDB之分片集群(Sharding)
MongoDB之分片集群(Sharding) 一.基本概念 分片(sharding)是一个通过多台机器分配数据的方法.MongoDB使用分片支持大数据集和高吞吐量的操作.大数据集和高吞吐量的数据库系统 ...
- LeetCode - 307. Range Sum Query - Mutable
Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive ...
- ECS的配置与使用
登录阿里云ECS,系统是centos7.2 在linux下通过useradd方式创建新用户,切换到该用户权限,发现-bash-4.2$ . 解决方法:先查看进程,关闭相关进程.然后使用useradd ...
- dnspython模块安装
wget http://www.dnspython.org/kits/1.12.0/dnspython-1.12.0.tar.gz tar -zxvf dnspython-1.12.0.tar.gz ...
- ipset批量配置iptables
简介: ipset是iptables的扩展,允许你创建匹配整个地址sets(地址集合)的规则.而不像普通的iptables链是线性的存储和过滤,ip集合存储在带索引的数据结构中,这种集合比较大也可以进 ...
- php获取今日开始时间戳和结束时间戳
1.php获取今日开始时间戳和结束时间戳 $beginToday=mktime(0,0,0,date('m'),date('d'),date('Y'));$endToday=mktime(0,0,0 ...
- iOS 9 HTTPS 的配置
方法有两种: (1)废话少说直接上图: (2)右击info.plist 文件 open as ->source code 在里面注入如下代码就行了(位置不固定,但要在指定的文件夹选项里) < ...
- 关于DOM与BOM的总结
1.什么是BOM,什么是DOM(基本概念) BOM: Browers Object MOdel 浏览器对象模型 DOM: Document Object MOdel ...
- C/C++语言简介之运算符
比较特别的是,比特右移(>>)运算符可以是算术(左端补最高有效位)或是逻辑(左端补 0)位移.例如,将 11100011 右移 3 比特,算术右移后成为 11111100,逻辑右移则为 0 ...
- ECharts 高度宽度自适应(转载)
最近在写一个地图类的应用,用的是echarts的图表,然而一上来就一脸懵逼,如果父级容器的height/width属性设置为百分比的形式,那么echarts就会warning,且不能正常的生成图表.所 ...