暴力算法解决几何问题-优快云博客

本文详细阐述了如何使用暴力算法解决几何问题，并着重介绍了vector容器的使用方法和erase函数的作用。通过实例展示了如何通过迭代器操作vector容器来优化算法效率。

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暴力题

注意vector用法，erase函数返回的是删除后的下一个元素的指针。迭代器的写法是：vector< >::iterator i;

View Code

#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <vector>
using namespace std;

#define maxn 100005
#define eps 1e-10

struct Point
{
double x, y;
};

struct Line
{
    Point a, b;
} line[maxn];

vector <int> stk(maxn);
int n;

void input()
{
for (int i = 0; i < n; i++)
        scanf("%lf%lf%lf%lf", &line[i].a.x, &line[i].a.y, &line[i].b.x, &line[i].b.y);
}

bool inter(Point &a, Point &b, Point &c, Point &d)
{
if (min(a.x, b.x) > max(c.x, d.x) || min(a.y, b.y) > max(c.y, d.y) || min(c.x, d.x) > max(a.x, b.x) || min(c.y, d.y) > max(a.y, b.y))
return 0;
double h = (b.x - a.x) * (c.y - a.y) - (b.y - a.y) * (c.x - a.x);
double i = (b.x - a.x) * (d.y - a.y) - (b.y - a.y) * (d.x - a.x);
double j = (d.x - c.x) * (a.y - c.y) - (d.y - c.y) * (a.x - c.x);
double k = (d.x - c.x) * (b.y - c.y) - (d.y - c.y) * (b.x - c.x);
return h * i <= eps && j * k <= eps;
}

bool cross(Line &p, Line &q)
{
return inter(p.a, p.b, q.a, q.b);
}

void make(Line &l)
{
for (vector<int>::iterator i = stk.begin(); i < stk.end();)
    {
if (cross(line[*i], l))
            i = stk.erase(i);
else
            i++;
    }
}

void work()
{
for (int i = 0; i < n; i++)
    {
        make(line[i]);
        stk.push_back(i);
    }
    printf("Top sticks: %d", stk[0] + 1);
for (vector<int>::iterator i = stk.begin() + 1; i < stk.end(); i++)
        printf(", %d", (*i) + 1);
    printf(".\n");
}

int main()
{
//freopen("t.txt", "r", stdin);
    while (scanf("%d", &n), n)
    {
        stk.clear();
        input();
        work();
    }
return 0;
}