题意:给出平面上的两类点,判断是否能画一条直线将两类点完全分割开来.

析:用暴力去枚举任意两点当作直线即可。

代码如下:

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e3 + 100;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
struct node{
int x, y, val;
bool operator < (const node &p) const{
return val < p.val;
}
};
node a[255]; int judge(const node &p1, const node &p2, const node &p3){
return (p1.x-p3.x) * (p2.y-p3.y) - (p1.y-p3.y) * (p2.x-p3.x);
}
bool cmp(const node &p, const node &q){
return p.x < q.x || (p.x == q.x && p.y < q.y);
} bool solve(int s, int t){
int x = 0, y = 0;
vector<node> v;
for(int i = 0; i < n; ++i){
if(i == s || t == i) continue;
if(!judge(a[s], a[t], a[i])){ v.push_back(a[i]); continue; }
if(judge(a[s], a[t], a[i]) > 0 && !a[i].val){
if(!x) x = 1, y = -1;
else if(x < 0) return false;
}
else if(judge(a[s], a[t], a[i]) < 0 && !a[i].val){
if(!x) x = -1, y = 1;
else if(x > 0) return false;
}
else if(judge(a[s], a[t], a[i]) > 0 && a[i].val){
if(!y) y = 1, x = -1;
else if(y < 0) return false;
}
else if(judge(a[s], a[t], a[i]) < 0 && a[i].val){
if(!y) y = -1, x = 1;
else if(y > 0) return false;
}
} if(!v.size()) return true;
int ok = 0;
v.push_back(a[s]);
v.push_back(a[t]);
sort(v.begin(), v.end(), cmp);
int cnt = 0;
for(int i = 0; i < v.size(); ++i){
if(v[i].val && !ok){
ok = 1;
}
else if(!v[i].val && !ok){
ok = -1;
}
else if(v[i].val && ok == -1){
ok = 1;
++cnt;
}
else if(!v[i].val && ok == 1){
ok = -1;
++cnt;
}
if(cnt > 1) return false;
}
return true;
} int main(){
int T; cin >> T;
while(T--){
scanf("%d", &n);
for(int i = 0; i < n; ++i){
scanf("%d %d %d", &a[i].x, &a[i].y, &a[i].val);
}
bool ok = false;
for(int i = 0; i < n; ++i){
for(int j = i+1; j < n; ++j)
if(solve(i, j)){ ok = true; break; }
if(ok) break;
} printf("%d\n", ok);
}
return 0;
}

UVaLive 7461 Separating Pebbles (暴力)的更多相关文章

  1. UVALive 7461 Separating Pebbles (计算几何)

    Separating Pebbles 题目链接: http://acm.hust.edu.cn/vjudge/contest/127401#problem/H Description http://7 ...

  2. UVALive7461 - Separating Pebbles 判断两个凸包相交

    //UVALive7461 - Separating Pebbles 判断两个凸包相交 #include <bits/stdc++.h> using namespace std; #def ...

  3. Gym 100299C && UVaLive 6582 Magical GCD (暴力+数论)

    题意:给出一个长度在 100 000 以内的正整数序列,大小不超过 10^ 12.求一个连续子序列,使得在所有的连续子序列中, 它们的GCD值乘以它们的长度最大. 析:暴力枚举右端点,然后在枚举左端点 ...

  4. UVALive 4423 String LD 暴力

    A - String LD Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Submit Stat ...

  5. UVaLive 3401 Colored Cubes (暴力)

    题意:给定n个立方体,让你重新涂尽量少的面,使得所有立方体都相同. 析:暴力求出每一种姿态,然后枚举每一种立方体的姿态,求出最少值. 代码如下: #pragma comment(linker, &qu ...

  6. UVALive 6912 Prime Switch 暴力枚举+贪心

    题目链接: https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show ...

  7. UVALive 6855 Banks (暴力)

    Banks 题目链接: http://acm.hust.edu.cn/vjudge/contest/130303#problem/A Description http://7xjob4.com1.z0 ...

  8. UVALive 7077 - Little Zu Chongzhi's Triangles(暴力)

    https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_probl ...

  9. UVaLive 6625 Diagrams & Tableaux (状压DP 或者 DFS暴力)

    题意:给一个的格子图,有 n 行单元格,每行有a[i]个格子,要求往格子中填1~m的数字,要求每个数字大于等于左边的数字,大于上边的数字,问有多少种填充方法. 析:感觉像个DP,但是不会啊...就想暴 ...

随机推荐

  1. Go --- GC优化经验

    不想看长篇大论的,这里先给个结论,go的gc还不完善但也不算不靠谱,关键看怎么用,尽量不要创建大量对象,也尽量不要频繁创建对象,这个道理其实在所有带gc的编程语言也都通用. 想知道如何提前预防和解决问 ...

  2. 安卓自己定义View进阶-Canvas之绘制基本形状

    Canvas之绘制基本形状 作者微博: @GcsSloop [本系列相关文章] 在上一篇自己定义View分类与流程中我们了解自己定义View相关的基本知识,只是,这些东西依然还是理论,并不能拿来(zh ...

  3. 【iOS】系统框架学习

    iOS的系统架构分为四个层次:核心操作系统层(Core OS layer).核心服务层(Core Services layer).媒体层(Media layer)和可触摸层(Cocoa Touch l ...

  4. Variable 'bop' is uninitialized when captured by block

    代码: - (void)doTest { NSBlockOperation * bop = [NSBlockOperation blockOperationWithBlock:^{ if (!bop. ...

  5. tabhost实现android菜单切换

    做APP项目已经有半个月了.慢慢地熟悉了这个开发环境和开发套路. 虽然是摸着石头过河.但也渐渐看到了水的深度! 作为一个电商项目APP,势必会涉及究竟部菜单条的功能.自己实现这个功能的过程是崎岖的,最 ...

  6. Spring Boot与Micronaut性能比较

    文章转载出处:微信公众号——锅外的大佬 链接:https://mp.weixin.qq.com/s/MdBByJ0ju-rROKg7jsWygA 今天我们将比较两个在JVM上构建微服务的框架:Spri ...

  7. Gym - 101164C - Castle KMP

    题目链接:传送门 题解: 利用KMP的fail失配数组,快速找到当前后缀与前缀的公共前缀点 #include<bits/stdc++.h> using namespace std; #pr ...

  8. HBase协处理器同步二级索引到Solr(续)

    一. 已知的问题和不足二.解决思路三.代码3.1 读取config文件内容3.2 封装SolrServer的获取方式3.3 编写提交数据到Solr的代码3.4 拦截HBase的Put和Delete操作 ...

  9. bash shell parameter expansion

    1 ${parameter%word}和${parameter%%word} ${parameter%word},word是一个模式,从parameter这个参数的末尾往前开始匹配.单个%进行最短匹配 ...

  10. SLG, 菱形格子的算法.(递归版

    class GeoPoint{ public: int x; int y; public: bool operator == (const GeoPoint& p){ return p.x = ...