Poj2749:Building roads
题意
有 N 个牛栏,现在通过一条通道(s1,s2)要么连到s1,要么连到s2,把他们连起来,他们之间有一些约束关系,一些牛栏不能连在同一个点,一些牛栏必须连在同一个点,现在问有没有可能把他们都连好,而且满足所有的约束关系,如果可以,输出距离最大的两个牛栏之间距离最小值(两点距离是指哈密顿距离)
Sol
二分答案+\(2-SAT\)判定
每次二分答案,把枚举两个点距离\(>mid\)的就连边限制
傻逼到没输出-1,WA无数遍
# include <iostream>
# include <stdio.h>
# include <stdlib.h>
# include <string.h>
# include <math.h>
# include <algorithm>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(1005);
const int __(4e6 + 5);
IL int Input(){
RG int x = 0, z = 1; RG char c = getchar();
for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
return x * z;
}
int n, A, B, tmp, first[_], cnt, num, mx, mn = 2e9;
int x[_], y[_], X1, Y1, X2, Y2, d1[_], d2[_], dis;
int S[_], vis[_], dfn[_], low[_], Index, col[_];
struct Link{
int u, v;
} hate[_], like[_];
struct Edge{
int to, next;
} edge[__];
IL void Add(RG int u, RG int v){
edge[cnt] = (Edge){v, first[u]}; first[u] = cnt++;
}
IL void Tarjan(RG int u){
vis[u] = 1, dfn[u] = low[u] = ++Index, S[++S[0]] = u;
for(RG int e = first[u]; e != -1; e = edge[e].next){
RG int v = edge[e].to;
if(!dfn[v]) Tarjan(v), low[u] = min(low[u], low[v]);
else if(vis[v]) low[u] = min(low[u], dfn[v]);
}
if(dfn[u] != low[u]) return;
RG int v = S[S[0]--]; col[v] = ++num, vis[v] = 0;
while(v != u) v = S[S[0]--], col[v] = num, vis[v] = 0;
}
IL int Calc(RG int a, RG int b, RG int c, RG int d){
return abs(a - c) + abs(b - d);
}
IL int Check(RG int mid){
Fill(first, -1), Fill(dfn, 0), Fill(col, 0), cnt = Index = num = 0;
for(RG int i = 1; i <= A; ++i){
Add(hate[i].u, hate[i].v + n), Add(hate[i].u + n, hate[i].v);
Add(hate[i].v, hate[i].u + n), Add(hate[i].v + n, hate[i].u);
}
for(RG int i = 1; i <= B; ++i){
Add(like[i].u, like[i].v), Add(like[i].u + n, like[i].v + n);
Add(like[i].v, like[i].u), Add(like[i].v + n, like[i].u + n);
}
for(RG int i = 1; i < n; ++i)
for(RG int j = i + 1; j <= n; ++j){
if(d1[i] + d1[j] > mid) Add(i, j + n), Add(j, i + n);
if(d2[i] + d2[j] > mid) Add(i + n, j), Add(j + n, i);
if(d1[i] + dis + d2[j] > mid) Add(i, j), Add(j + n, i + n);
if(d2[i] + dis + d1[j] > mid) Add(i + n, j + n), Add(j, i);
}
for(RG int tmp = n << 1, i = 1; i <= tmp; ++i) if(!dfn[i]) Tarjan(i);
for(RG int i = 1; i <= n; ++i) if(col[i] == col[i + n]) return 0;
return 1;
}
int main(RG int argc, RG char* argv[]){
n = Input(), A = Input(), B = Input();
X1 = Input(), Y1 = Input(), X2 = Input(), Y2 = Input();
dis = Calc(X1, Y1, X2, Y2);
for(RG int i = 1; i <= n; ++i){
x[i] = Input(), y[i] = Input();
d1[i] = Calc(x[i], y[i], X1, Y1);
d2[i] = Calc(x[i], y[i], X2, Y2);
mx = max(mx, max(d1[i], d2[i]));
mn = min(mn, min(d1[i], d2[i]));
}
for(RG int i = 1; i <= A; ++i) hate[i] = (Link){Input(), Input()};
for(RG int i = 1; i <= B; ++i) like[i] = (Link){Input(), Input()};
mx = mx * 2 + dis, mn *= 2;
RG int l = mn, r = mx, ans = -1;
while(l <= r){
RG int mid = (l + r) >> 1;
if(Check(mid)) ans = mid, r = mid - 1;
else l = mid + 1;
}
printf("%d\n", ans);
return 0;
}
Poj2749:Building roads的更多相关文章
- POJ2749:Building roads——题解
http://poj.org/problem?id=2749 (这个约翰的奶牛真多事…………………………) i表示u与s1连,i+n表示u与s2连. 老规矩,u到v表示取u必须取v. 那么对于互相打架 ...
- [POJ2749]Building roads(2-SAT)
Building roads Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 8153 Accepted: 2772 De ...
- 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 1815, POJ 2749 Building roads(2-sat)
HDU 1815, POJ 2749 Building roads pid=1815" target="_blank" style="">题目链 ...
- Building roads
Building roads Time Limit: 10000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Tot ...
- bzoj 1626: [Usaco2007 Dec]Building Roads 修建道路 -- 最小生成树
1626: [Usaco2007 Dec]Building Roads 修建道路 Time Limit: 5 Sec Memory Limit: 64 MB Description Farmer J ...
- 洛谷——P2872 [USACO07DEC]道路建设Building Roads
P2872 [USACO07DEC]道路建设Building Roads 题目描述 Farmer John had just acquired several new farms! He wants ...
- USACO Building Roads
洛谷 P2872 [USACO07DEC]道路建设Building Roads 洛谷传送门 JDOJ 2546: USACO 2007 Dec Silver 2.Building Roads JDOJ ...
随机推荐
- PLECS—直流电机系统2
1.模型图 2,计算及仿真 1)计算 2)仿真 n = 1870.1 r/min (wm = 195.833 rad/s) ...
- css实现隐藏多余溢出文字并显示省略号
<meta charset="utf-8" /> <style> .txt{ width:200px; border:1px solid #ddd; ove ...
- 【NOIP2012】 疫情控制
[NOIP2012] 疫情控制 标签: 倍增 贪心 二分答案 NOIP Description H 国有 n 个城市,这 n 个城市用 n-1 条双向道路相互连通构成一棵树, 1 号城市是首都, 也是 ...
- 微信小程序下拉框
微信小程序里没有和HTML里的下拉框一样的组件,想要相同的效果只能自己写一个,先看效果 下面来看一下代码: 首先WXML <view class='select_box'> <vie ...
- js获取某个日期所在周周一的日期
第一次写,做个小笔记. 第一步:获取该日期的星期数: 第二步:在该日期上减去他的星期数再减1,(注:星期日获取到的星期数是0): 下面是具体代码: function GetMonday(dd) { v ...
- django新手第一课
django是基于python的一个web框架,大致结构如下: 在pycharm,python2.7,django1.8,mysql都装好的情况下,现在开始django的初试: 一.基础启动djang ...
- 沉淀,再出发——安装windows10和ubuntu kylin15.04双系统心得体会
安装windows10和ubuntu kylin15.04双系统心得体会 一.安装次序 很简单,两种安装次序,"先安装windows后安装linux:先安装linux后安装wind ...
- Storm日志分析调研及其实时架构
1.Storm第一个Demo 2.Windows下基于eclipse的Storm应用开发与调试 3.Storm实例+mysql数据库保存 4.Storm原理介绍 5. flume+kafka+stor ...
- java-数据库连接,分层实现增删改查测试
成员属性类: public class Dog { private int number; private String name; private String strain; private St ...
- linux dns搭建
DNS:域名解析(Domain Nmae System)正向解析:根据主机名称(域名)查找其对应的ip地址,这是最基本,最常用的功能反向解析:根据ip地址查找其对应的主机名称(域名),反垃圾邮件/安全 ...