bzoj1091[SCOI2003]切割多边形

【算法】

【注意】

【代码】

/**************************************************************
Problem: 1091
User: zhengly123
Language: C++
Result: Accepted
Time:124 ms
Memory:1276 kb
****************************************************************/

#include <stdlib.h>
#include <stdio.h>
#include <iostream>
#include <math.h>
#include <algorithm>
#include <string.h>
#define MEM(a) memset(a,0,sizeof(a))
#define CPY(a,b) memcpy(b,a,sizeof(b))
using namespace std;
const int INF = 999999999;
const int MAXN = 20;
const double eps = 1e-8;

int zero(double a)
{
if (fabs(a) < eps) return 0;
else return (a>0) ? 1 : -1;
}
double MIN(double a, double b){ return (a < b) ? a : b; }

struct point//点或向量
{
double x, y;
point(){}
point(double a, double b){ x = a; y = b; }
friend point operator+(const point&a, const point&b){ return point(a.x + b.x, a.y + b.y); }
friend point operator -(const point&a, const point&b){ return point(a.x - b.x, a.y - b.y); }
friend point operator *(const double&k, const point&a){ return point(k*a.x, k*a.y); }//数乘
friend double operator *(const point&a, const point&b){ return a.x*b.x + a.y*b.y; }//点积
friend double operator /(const point&a, const point&b){ return a.x*b.y - a.y*b.x; }//叉积,用^优先级有问题
friend bool operator <(const point&a, const point&b){ return (a.x - b.x < -eps) || (fabs(a.x - b.x) < eps && (a.y - b.y) < -eps); }
friend bool operator ==(const point&a, const point&b){ return zero(a.x - b.x) && zero(a.y - b.y); }
double dis(){ return (sqrt(x*x + y*y)); }
void input(){ scanf("%lf%lf", &x, &y); }
void output(){ printf("%lf %lf\n", x, y); }
}p[MAXN], pt[MAXN];

point inter(point a, point b, point c, point d)
{
double s1 = (b - a) / (d - a), s2 = (b - a) / (c - a);
return (point((s1*c.x - s2*d.x) / (s1 - s2), (s1*c.y - s2*d.y) / (s1 - s2)));
}

int nump;
double ans = INF;

int cut(point pp[], int num,int k, double &dis)//the number of p is num, cut the (k,k+1)半平面交
{
int i, j, cnt = 0, newcnt = -1, t;
point newp[2];
MEM(pt);
for (i = 1; i <= num; ++i)
{
t = zero((p[k + 1] - p[k]) / (pp[i] - p[k]));
if (t >= 0)
{
pt[++cnt] = pp[i];
if (t == 0) newp[++newcnt] = pp[i];
}
else
{
if (zero((p[k + 1] - p[k]) / (pp[i - 1] - p[k])) > 0)
newp[++newcnt] = pt[++cnt] = inter(p[k], p[k + 1], pp[i], pp[i - 1]);
if (zero((p[k + 1] - p[k]) / (pp[i + 1] - p[k])) > 0)
newp[++newcnt] = pt[++cnt] = inter(p[k], p[k + 1], pp[i], pp[i + 1]);
}
}
pt[0] = pt[cnt]; pt[cnt + 1] = pt[1];
dis += (newp[1] - newp[0]).dis();
return cnt;
}

void dfs(int dep, point p[], int num, bool boo[],double dis)
{
int curnum;
double curdis;
point curp[MAXN];
bool curboo[MAXN];
if (dep == nump + 1)
{
ans = MIN(ans, dis);
return;
}
CPY(boo, curboo);
for (int i = 1; i <= nump; ++i)
{
if (!boo[i])
{
CPY(p, curp);
curboo[i] = 1;
curdis = dis;
curnum = cut(curp, num, i, curdis);
CPY(pt, curp);
dfs(dep + 1, curp, curnum, curboo, curdis);
curboo[i] = 0;
}
}
}

void work()
{
int i, j, n, m;
bool boo[MAXN] = {};
point pt[MAXN];
scanf("%d%d%d", &n, &m, &nump);
for (i = nump; i >= 0; --i) p[i].input();
p[0] = p[nump]; p[nump + 1] = p[1];
pt[1] = pt[5] = point(0, 0);
pt[2] = point(n, 0);
pt[3] = point(n, m);
pt[4] = pt[0] = point(0, m);
dfs(1, pt, 4, boo, 0);
printf("%.3lf\n", ans);
}

int main()
{
work();
return 0;
}