UVaLive 7461 Separating Pebbles (暴力)

Posted 2020-08-09 dwtfukgv

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;
}