bzoj4153 [Ipsc2015]Familiar Couples
Description
Input
Output
#include<cstdio>#include<vector>
const int P=1e9+,N=1e6+;
inline int read(){
int x=,c=getchar();
while(c>''||c<'')c=getchar();
while(c>=''&&c<='')x=x*+c-'',c=getchar();
return x;
}
inline void exch(int&a,int&b){int c=a;a=b;b=c;}namespace Map{
const int mx=;
unsigned int nw=;
unsigned int xs[mx],ys[mx],zs[mx],ds[mx];
inline void clear(){
nw++;
}
inline int get(unsigned int x,unsigned int y,int inc){
unsigned int w=(x*+y*+)%mx;
while(ds[w]==nw){
if(xs[w]==x&&ys[w]==y){
int v=zs[w];
zs[w]+=inc;
return v;
}
w+=;
if(w>=mx)w-=mx;
}
ds[w]=nw;
xs[w]=x;ys[w]=y;zs[w]=inc;
return ;
}
}
int T,n,m,now=,Ans,ans;
int h1[N],h2[N],f1[N],f2[N],sz1[N],sz2[N],nx1[N],nx2[N];
int t[N],d[N];
std::vector<int>v1[N],v2[N];
int main(){
T=read();
while(T--){
n=read();m=read();
Ans=ans=;
Map::clear();
for(int i=;i<=n;i++){
Map::get(i,i,);
f1[i]=f2[i]=i;
v1[i].clear();v2[i].clear();
v1[i].push_back(i);
v2[i].push_back(i);
}
for(int i=,op,a,b;i<=m;++i){
op=read();a=read();b=read();
++now;
if(op==){
if(f1[a]!=f1[b]){
if(v1[f1[a]].size()>v1[f1[b]].size())exch(a,b);
std::vector<int>&vc=v1[f1[a]];
for(int x=;x<vc.size();x++){
int p=vc[x];
int f=f2[p];
if(d[f]!=now)d[f]=now,ans=(ans+Map::get(f1[p],f,-)*1ll*Map::get(f1[b],f,)%P)%P;
else Map::get(f1[p],f,-),Map::get(f1[b],f,);
f1[p]=f1[b];
v1[f1[b]].push_back(p);
}
vc.clear();
}
}else{
if(f2[a]!=f2[b]){
if(v2[f2[a]].size()>v2[f2[b]].size())exch(a,b);
std::vector<int>&vc=v2[f2[a]];
for(int x=;x<vc.size();x++){
int p=vc[x];
int f=f1[p];
if(d[f]!=now)d[f]=now,ans=(ans+Map::get(f,f2[p],-)*1ll*Map::get(f,f2[b],)%P)%P;
else Map::get(f,f2[p],-),Map::get(f,f2[b],);
f2[p]=f2[b];
v2[f2[b]].push_back(p);
}
vc.clear();
}
}
Ans=(Ans+ans*1ll*i%P)%P;
}
printf("%d\n",Ans);
}
return ;
}
bzoj4153 [Ipsc2015]Familiar Couples的更多相关文章
- bzoj AC倒序
Search GO 说明:输入题号直接进入相应题目,如需搜索含数字的题目,请在关键词前加单引号 Problem ID Title Source AC Submit Y 1000 A+B Problem ...
- Sicily 1021. Couples
题目地址:1021. Couples 思路: 想清楚了这道题其实很简单.利用夫妻出现的位置作为下标,并设为同一值,第一对夫妻值为1,第二对为2,以此类推,存储完毕即可进入下一步. 利用栈这个数据结构: ...
- Julia is a high-level, high-performance dynamic programming language for technical computing, with syntax that is familiar to users of other technical
http://julialang.org/ julia | source | downloads | docs | blog | community | teaching | publications ...
- [LeetCode] Couples Holding Hands 两两握手
N couples sit in 2N seats arranged in a row and want to hold hands. We want to know the minimum numb ...
- [Swift]LeetCode765. 情侣牵手 | Couples Holding Hands
N couples sit in 2N seats arranged in a row and want to hold hands. We want to know the minimum numb ...
- [CC-COUPLES]Couples sit next to each other
[CC-COUPLES]Couples sit next to each other 题目大意: 有\(n(n\le5\times10^5)\)对小伙伴共\(2n\)个人坐成一圈.刚开始编号为\(i\ ...
- 每日英语:Why Rate Your Marriage? A Numerical Score Can Help Couples Talk About Problems
When marriage therapist Sharon Gilchrest O'Neill met with new clients recently, she asked them why t ...
- ZOJ 3161 Damn Couples 动态规划 难度:2
Damn Couples Time Limit: 1 Second Memory Limit: 32768 KB As mentioned in the problem "Coup ...
- LeetCode765. Couples Holding Hands
N couples sit in 2N seats arranged in a row and want to hold hands. We want to know the minimum numb ...
随机推荐
- 以太网最大帧和最小帧、MTU
根据rfc894的说明,以太网封装IP数据包的最大长度是1500字节,也就是说以太网最大帧长应该是以太网首部加上1500,再加上7字节的前导同步码和1字节的帧开始定界符,具体就是:7字节前导同步码 + ...
- SharePoint 2010 Ribbon with wrong style in Chrome and Safari
When we add custom ribbon to SharePoint 2010, it may display well in IE but not in Chrome and Safari ...
- 第一次做Java程序注意事项
public class myapp{ public static void main(String[] args){ System.out.println("Hallo Java!&quo ...
- 玩转TypeScript(3)--数组
数组的语法和语义与C#数组非常相似,首先要指定一个数组名,后跟冒号,冒号后面紧跟数组的类型,数组类型名后面跟方括号表示当前定义的是一个数组,对于一个空的数组定义,可以使用如下的语法: btnArray ...
- Linux C socket 封装
/************************************************************************** * Linux C socket 封装 * 声明 ...
- OK335xS-Android mkmmc-android-ubifs.sh hacking
#/******************************************************************************* # * OK335xS-Androi ...
- 20155216 2016-2017-2 《Java程序设计》第五周学习总结
20155216 2016-2017-2 <Java程序设计>第五周学习总结 教材学习内容总结 使用try,catch,finally处理异常 JVM会尝试执行try区块中的程序代码,如果 ...
- Linq 增删改查
数据库访问技术: ADO.net EF框架 LinQ LinQ是一种高集成化的数据库访问技术,他将数据库中的表映射成程序中的类 数据库的表名变成类名 数据库的列名变成字段名/属性名 所有的操作都是通过 ...
- java泛型学习(2)
一:深入泛型使用.主要是父类和子类存在泛型的demo /** * 父类为泛型类 * @author 尚晓飞 * @date 2014-7-15 下午7:31:25 * * * 父类和子类的泛型. * ...
- 2017年最新cocoapods安装教程(解决淘宝镜像源无效以及其他源下载慢问题)
首先,先来说一下一般的方法吧,就是把之前的淘宝源替换成一个可用的的源: 使用终端查看当前的源 gem sources -l gem sources -r https://rubygems.org/ # ...