HDU 1815 Building roads
二分答案 + 2-SAT验证
POJ 稳过,HDU C++ 超时,G++ 550ms左右AC
#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<stack>
#include<algorithm>
using namespace std; const int maxn=+;
int N,A,B;
int left,right,mid;
int s1x,s1y,s2x,s2y;
struct point
{
int x,y;
int p_s1;
int p_s2;
}p[maxn];
int Hx[maxn],Hy[maxn];
int Lx[maxn],Ly[maxn]; stack<int>S;
vector<int>G[maxn];
vector<int>FG[maxn];
int Belong[maxn];
int flag[maxn];
int Block; int dis1,dis2,dis3; void init()
{
for(int i=;i<maxn;i++) G[i].clear();
for(int i=;i<maxn;i++) FG[i].clear();
memset(Belong,,sizeof Belong);
memset(flag,,sizeof flag);
while(!S.empty()) S.pop();
Block=;
} void addEgde(int x,int y)
{
G[x].push_back(y);
FG[y].push_back(x);
} int Dis(int x1,int y1,int x2,int y2)
{
return abs(x1-x2)+abs(y1-y2);
} void read()
{
scanf("%d%d%d%d",&s1x,&s1y,&s2x,&s2y);
for(int i=;i<N;i++)
scanf("%d%d",&p[i].x,&p[i].y);
for(int i=;i<=A;i++)
{
scanf("%d%d",&Hx[i],&Hy[i]);
Hx[i]--;Hy[i]--;
}
for(int i=;i<=B;i++)
{
scanf("%d%d",&Lx[i],&Ly[i]);
Lx[i]--;Ly[i]--;
}
} void dfs1(int now)
{
flag[now]=;
for(int i=;i<G[now].size();i++)
if(!flag[G[now][i]])
dfs1(G[now][i]);
S.push(now);
} void dfs2(int now)
{
Belong[now]=Block;
for(int i=;i<FG[now].size();i++)
if(!Belong[FG[now][i]])
dfs2(FG[now][i]);
} bool judge()
{
for(int i=;i<*N;i++) if(!flag[i]) dfs1(i);
while(!S.empty())
{
int Top=S.top(); S.pop();
if(!Belong[Top])
{
Block++;
dfs2(Top);
}
}
for(int i=;i<N;i++)
if(Belong[*i]==Belong[*i+])
return ;
return ;
} void solve()
{
left=;right=;
int ans=-;
while(left<=right)
{
mid=(left+right)/;
init(); for(int i=;i<N;i++){
for(int j=i+;j<N;j++)
{
int pi_s1=p[i].p_s1;
int pi_s2=p[i].p_s2;
int pj_s1=p[j].p_s1;
int pj_s2=p[j].p_s2;
int s1_s2=Dis(s1x,s1y,s2x,s2y); if(pi_s1+pj_s1>mid) {
addEgde(*i,*j+);
addEgde(*j,*i+);
} if(pi_s2+pj_s2>mid) {
addEgde(*i+,*j);
addEgde(*j+,*i);
} if(pi_s1+s1_s2+pj_s2>mid) {
addEgde(*i,*j);
addEgde(*j+,*i+);
} if(pi_s2+s1_s2+pj_s1>mid) {
addEgde(*i+,*j+);
addEgde(*j,*i);
}
} } for(int i=;i<=A;i++)
{
addEgde(*Hx[i],*Hy[i]+);
addEgde(*Hy[i]+,*Hx[i]);
addEgde(*Hx[i]+,*Hy[i]);
addEgde(*Hy[i],*Hx[i]+);
}
for(int i=;i<=B;i++)
{
addEgde(*Lx[i],*Ly[i]);
addEgde(*Ly[i],*Lx[i]);
addEgde(*Lx[i]+,*Ly[i]+);
addEgde(*Ly[i]+,*Lx[i]+);
}
if(judge()) ans=mid,right=mid-;
else left=mid+;
}
printf("%d\n",ans);
}
int main()
{
while(~scanf("%d%d%d",&N,&A,&B))
{
read();
for(int i=;i<N;i++)
{
p[i].p_s1=Dis(p[i].x,p[i].y,s1x,s1y);
p[i].p_s2=Dis(p[i].x,p[i].y,s2x,s2y);
}
solve();
}
return ;
}
HDU 1815 Building roads的更多相关文章
- HDU 3721 Building Roads (2010 Asia Tianjin Regional Contest) - from lanshui_Yang
感慨一下,区域赛的题目果然很费脑啊!!不过确实是一道不可多得的好题目!! 题目大意:给你一棵有n个节点的树,让你移动树中一条边的位置,即将这条边连接到任意两个顶点(边的大小不变),要求使得到的新树的直 ...
- HDU 1815, POJ 2749 Building roads(2-sat)
HDU 1815, POJ 2749 Building roads pid=1815" target="_blank" style="">题目链 ...
- poj 3625 Building Roads
题目连接 http://poj.org/problem?id=3625 Building Roads Description Farmer John had just acquired several ...
- poj 2749 Building roads (二分+拆点+2-sat)
Building roads Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 6229 Accepted: 2093 De ...
- BZOJ 1626: [Usaco2007 Dec]Building Roads 修建道路( MST )
计算距离时平方爆了int结果就WA了一次...... ------------------------------------------------------------------------- ...
- HDU 1102 Constructing Roads, Prim+优先队列
题目链接:HDU 1102 Constructing Roads Constructing Roads Problem Description There are N villages, which ...
- HDU 5033 Building(单调栈)
HDU 5033 Building(单调栈) 题目链接http://acm.hdu.edu.cn/showproblem.php?pid=5033 Description Once upon a ti ...
- Building roads
Building roads Time Limit: 10000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Tot ...
- HDU—— 5159 Building Blocks
Problem Description After enjoying the movie,LeLe went home alone. LeLe decided to build blocks. LeL ...
随机推荐
- js 学习总结
new array()[] []表示数组new object(){} {}表示对象 JavaScript 对象 对象由花括号分隔.在括号内部,对象的属性以名称和值对的形式 (name : value) ...
- Codeforces Round #364 (Div. 2) D. As Fast As Possible
D. As Fast As Possible time limit per test 1 second memory limit per test 256 megabytes input stand ...
- ios根据颜色返回图片
+(UIImage*) createImageWithColor:(UIColor*) color { CGRect rect=CGRectMake(0.0f, 0.0f, 1.0f, 1.0f); ...
- Ansible7:Playbook常用模块【转】
playbook的模块与在ansible命令行下使用的模块有一些不同.这主要是因为在playbook中会使用到一些facts变量和一些通过setup模块从远程主机上获取到的变量.有些模块没法在命令行下 ...
- 近十年one-to-one最短路算法研究整理【转】
前言:针对单源最短路算法,目前最经典的思路即标号算法,以Dijkstra算法和Bellman-Ford算法为根本演进了各种优化技术和算法.针对复杂网络,传统的优化思路是在数据结构和双向搜索上做文章,或 ...
- rsync unison+inotify双向实时同步
rsync多线程同步 A:文件服务器 ip:10.10.1.10 B:备份服务器 ip:10.10.1.11 1.在B服务器上安装rsync软件 tar xzvf rsync-3.1.0.tar.gz ...
- zookeeper集群实例
zookeeper是什么 Zookeeper,一种分布式应用的协作服务,是Google的Chubby一个开源的实现,是Hadoop的分布式协调服务,它包含一个简单的原语集,应用于分布式应用的协作服务, ...
- Django 分页功能
Django 分页功能比较强大,这边是结合官网的内容写的可以参考 https://docs.djangoproject.com/en/1.9/topics/pagination/ 分页命令行练习案列 ...
- linux sigaction信号处理
sigaction函数相比signal函数更为复杂,但更具灵活性,下面具体介绍她的结构和用法: #include <signal.h> int sigaction(int signum, ...
- Windsock套接字I/O模型学习 --- 第三章
1. WSAAsyncSelect 模型 WSAAsyncSelect 模型比较简单,是为了适应Windows的消息驱动环境而设置的,WSAAsyncSelect 函数自动把套接字设为非阻塞模式.MF ...