poj 1271 || UVA 10117 Nice Milk

纠结了一下午啊！！！超时啊！！！我用WS的方法测了好久又测了下我的时间复杂度！！！感觉应该就卡在数据的最后一组啊！！！！

在UVA wa了啊！！！说明UVA 机器比POJ快啊！！！

思路：求蘸牛奶的最大面积，也就是半平面交后中间区域最小，所以只需要求得半平面交中间的最小区域即可。这个实现需要DFS枚举哈。

一直TLE，后来看网上的还是不知道从哪入手。后来想起一种优化方法，因为里面的边是平行的，所以如果选择里面的边，那么外面的那个边就不需要了（因为会在排序后删掉的），那么直接覆盖那条边就行了，还省点时间。这么一弄，居然WA了 = = 后来发现个很2的错误，输入放在输出前了 T T 。。。输出0.00后直接continue了 = = 改掉后AC。

两个代码都是DFS的差异，一个是查找到一条边就加入然后到k条后求面积。另一种是把DFS方案全部列出来，然后再纠结 = =。。。

第一种快点，因为如果找到面积为0的我就直接return了。而且，如果剩下可选的元素比需要选的元素个数少，也return了。不过最少还600+MS.。。。T T.。。

P.S. 我怀疑过unique的删除，因为如果用半平面交的话，第一次删除斜率相等的点的话，应该留下最左边的（我定义最左边为有效方向）。那么我怕unique乱删。不过刚才测试了下，unique删的都是第一个元素，那么我就放心了。

P.S. 刚查了下unique的内部实现，确实是保留第一个元素，呵呵。

#include <cmath>#include <cstdio>#include <cstdlib>#include <iostream>#include <cstring>#include <string>#include <algorithm>#include <queue>#include <stack>#include <climits>using namespace std;const int MAX = 25;const double inf = 1e20;struct point {double x,y;};struct line { double ang; point a,b;};point p[MAX],s[MAX];line ln[MAX],deq[MAX],ltmp[MAX],ll[MAX],lk[MAX];;double h,mmin;const double eps = 1e-6;int k;bool dy(double x,double y){return x > y + eps;}// x > y bool xy(double x,double y){return x < y - eps;}// x < y bool dyd(double x,double y){ return x > y - eps;}// x >= y bool xyd(double x,double y){return x < y + eps;} // x <= y bool dd(double x,double y) {return fabs( x - y ) < eps;}  // x == ydouble disp2p(point a,point b) //  a b 两点之间的距离 {return sqrt( ( a.x - b.x ) * ( a.x - b.x ) + ( a.y - b.y ) * ( a.y - b.y ) );}double crossProduct(point a,point b,point c)//向量 ac 在 ab 的方向 {return (c.x - a.x)*(b.y - a.y) - (b.x - a.x)*(c.y - a.y);}void changepoint(point a,point b,point &c,point &d){double len = disp2p(a,b);    double dx = h / len * ( a.y - b.y );double dy = h / len * (-a.x + b.x );c.x = a.x + dx; c.y = a.y + dy;d.x = b.x + dx; d.y = b.y + dy;}void makeline_hp(point a,point b,line &l) // 线段(向量ab)右侧区域有效 {l.a = a; l.b = b;l.ang = atan2(a.y - b.y,a.x - b.x);// 如果是左侧区域，改成b.y - a.y,b.x - a.x }bool equal_ang(line a,line b)// 第一次unique的比较函数 {return dd(a.ang,b.ang);}bool cmphp(line a,line b)// 排序的比较函数 {if( dd(a.ang,b.ang) ) return xy(crossProduct(b.a,b.b,a.a),0.0);return xy(a.ang,b.ang);}bool equal_p(point a,point b)//第二次unique的比较函数 {return dd(a.x,b.x) && dd(a.y,b.y);}bool parallel(line u,line v){return dd( (u.a.x - u.b.x)*(v.a.y - v.b.y) - (v.a.x - v.b.x)*(u.a.y - u.b.y) , 0.0 );}point l2l_inst_p(line l1,line l2){point ans = l1.a;double t = ((l1.a.x - l2.a.x)*(l2.a.y - l2.b.y) - (l1.a.y - l2.a.y)*(l2.a.x - l2.b.x))/   ((l1.a.x - l1.b.x)*(l2.a.y - l2.b.y) - (l1.a.y - l1.b.y)*(l2.a.x - l2.b.x));ans.x += (l1.b.x - l1.a.x)*t;ans.y += (l1.b.y - l1.a.y)*t;return ans;}double area_polygon(point p[],int n){double s = 0.0;for(int i=0; i<n; i++)s += p[(i+1)%n].y * p[i].x - p[(i+1)%n].x * p[i].y;return fabs(s)/2.0;}void inst_hp_nlogn(line *ln,int n,point *s,int &len){len = 0;sort(ln,ln+n,cmphp);n = unique(ln,ln+n,equal_ang) - ln;int bot = 0,top = 1;deq[0] = ln[0]; deq[1] = ln[1];for(int i=2; i<n; i++){if( parallel(deq[top],deq[top-1]) || parallel(deq[bot],deq[bot+1]) )return ;while( bot < top && dy(crossProduct(ln[i].a,ln[i].b,l2l_inst_p(deq[top],deq[top-1])),0.0) )top--;while( bot < top && dy(crossProduct(ln[i].a,ln[i].b,l2l_inst_p(deq[bot],deq[bot+1])),0.0) )bot++;deq[++top] = ln[i];}while( bot < top && dy(crossProduct(deq[bot].a,deq[bot].b,l2l_inst_p(deq[top],deq[top-1])),0.0) )top--;while( bot < top && dy(crossProduct(deq[top].a,deq[top].b,l2l_inst_p(deq[bot],deq[bot+1])),0.0) )bot++;if( top <= bot + 1 ) return ;for(int i=bot; i<top; i++)s[len++] = l2l_inst_p(deq[i],deq[i+1]);if( bot < top + 1 ) s[len++] = l2l_inst_p(deq[bot],deq[top]);len = unique(s,s+len,equal_p) - s;}void DFS(int x,int n,int sum){if( mmin == 0.0 ) return ;if( n - x < k - sum ) return ;if( sum == k ){int len;memcpy(ltmp,ln,sizeof(ln));inst_hp_nlogn(ltmp,n,s,len);mmin = min(mmin,area_polygon(s,len));if( mmin == 0.0 ) return ;return ;}for(int i=x; i<n; i++){ln[i] = ll[i];DFS(i+1,n,sum+1);if( mmin == 0.0 ) return ;ln[i] = lk[i];}}int main(){int n;while( ~scanf("%d%d%lf",&n,&k,&h) ){if( !(n || k || h) ) break;for(int i=0; i<n; i++)scanf("%lf%lf",&p[i].x,&p[i].y);if( h == 0 || k == 0 ){printf("0.00\n");continue;}p[n] = p[0];double area = area_polygon(p,n);for(int i=0; i<n; i++){point c,d;changepoint(p[i],p[i+1],c,d);makeline_hp(c,d,ll[i]);makeline_hp(p[i],p[i+1],ln[i]);}memcpy(lk,ln,sizeof(ln));if( k > n ) k = n;mmin = inf;DFS(0,n,0);double ans = area - mmin;printf("%.2lf\n",ans);}return 0;}

#include <cmath>#include <cstdio>#include <cstdlib>#include <iostream>#include <cstring>#include <string>#include <algorithm>#include <queue>#include <stack>#include <climits>using namespace std;const int MAX = 25;const double inf = 1e20;const double pi = acos(-1.0);struct point {double x,y;};struct line { double ang; point a,b;};point p[MAX],s[MAX];line ln[MAX],deq[MAX],ltmp[MAX],ll[MAX];double h,mmin;const double eps = 1e-6;int k,cnt;bool dy(double x,double y){return x > y + eps;}// x > y bool xy(double x,double y){return x < y - eps;}// x < y bool dyd(double x,double y){ return x > y - eps;}// x >= y bool xyd(double x,double y){return x < y + eps;} // x <= y bool dd(double x,double y) {return fabs( x - y ) < eps;}  // x == ydouble disp2p(point a,point b) //  a b 两点之间的距离 {return sqrt( ( a.x - b.x ) * ( a.x - b.x ) + ( a.y - b.y ) * ( a.y - b.y ) );}double crossProduct(point a,point b,point c)//向量 ac 在 ab 的方向 {return (c.x - a.x)*(b.y - a.y) - (b.x - a.x)*(c.y - a.y);}void changepoint(point a,point b,point &c,point &d){double len = disp2p(a,b);    double dx = h / len * ( a.y - b.y );double dy = h / len * (-a.x + b.x );c.x = a.x + dx; c.y = a.y + dy;d.x = b.x + dx; d.y = b.y + dy;}void makeline_hp(point a,point b,line &l){l.a = a; l.b = b;l.ang = atan2(a.y - b.y,a.x - b.x);}bool equal_ang(line a,line b)// 第一次unique的比较函数 {return dd(a.ang,b.ang);}bool cmphp(line a,line b)// 排序的比较函数 {if( dd(a.ang,b.ang) ) return xy(crossProduct(b.a,b.b,a.a),0.0);return xy(a.ang,b.ang);}bool equal_p(point a,point b)//第二次unique的比较函数 {return dd(a.x,b.x) && dd(a.y,b.y);}bool parallel(line u,line v){return dd( (u.a.x - u.b.x)*(v.a.y - v.b.y) - (v.a.x - v.b.x)*(u.a.y - u.b.y) , 0.0 );}point l2l_inst_p(line l1,line l2){point ans = l1.a;double t = ((l1.a.x - l2.a.x)*(l2.a.y - l2.b.y) - (l1.a.y - l2.a.y)*(l2.a.x - l2.b.x))/   ((l1.a.x - l1.b.x)*(l2.a.y - l2.b.y) - (l1.a.y - l1.b.y)*(l2.a.x - l2.b.x));ans.x += (l1.b.x - l1.a.x)*t;ans.y += (l1.b.y - l1.a.y)*t;return ans;}double area_polygon(point p[],int n){if( n < 3 ) return 0.0;double s = 0.0;for(int i=0; i<n; i++)s += p[(i+1)%n].y * p[i].x - p[(i+1)%n].x * p[i].y;return fabs(s)/2.0;}line lk[MAX];void inst_hp_nlogn(line *ln,int n,point *s,int &len){len = 0;sort(ln,ln+n,cmphp);n = unique(ln,ln+n,equal_ang) - ln;int bot = 0,top = 1;deq[0] = ln[0]; deq[1] = ln[1];for(int i=2; i<n; i++){if( parallel(deq[top],deq[top-1]) || parallel(deq[bot],deq[bot+1]) )return ;while( bot < top && dy(crossProduct(ln[i].a,ln[i].b,l2l_inst_p(deq[top],deq[top-1])),0.0) )top--;while( bot < top && dy(crossProduct(ln[i].a,ln[i].b,l2l_inst_p(deq[bot],deq[bot+1])),0.0) )bot++;deq[++top] = ln[i];}while( bot < top && dy(crossProduct(deq[bot].a,deq[bot].b,l2l_inst_p(deq[top],deq[top-1])),0.0) )top--;while( bot < top && dy(crossProduct(deq[top].a,deq[top].b,l2l_inst_p(deq[bot],deq[bot+1])),0.0) )bot++;if( top <= bot + 1 ) return ;for(int i=bot; i<top; i++)s[len++] = l2l_inst_p(deq[i],deq[i+1]);if( bot < top + 1 ) s[len++] = l2l_inst_p(deq[bot],deq[top]);len = unique(s,s+len,equal_p) - s;}int t[150000][10];int tt[10];void DFS(int x,int n,int sum){if( n - x < k - sum ) return ;if( sum == k ){memcpy(t[cnt++],tt,sizeof(tt));return ;}for(int i=x; i<n; i++){tt[sum] = i;DFS(i+1,n,sum+1);}}int main(){int n,len;point c,d;while( ~scanf("%d%d%lf",&n,&k,&h) ){if( !(n || k || h) ) break;cnt = 0;for(int i=0; i<n; i++)scanf("%lf%lf",&p[i].x,&p[i].y);if( h == 0 || k == 0 ){printf("0.00\n");continue;}p[n] = p[0];double area = area_polygon(p,n);for(int i=0; i<n; i++){changepoint(p[i],p[i+1],c,d);makeline_hp(c,d,ll[i]);makeline_hp(p[i],p[i+1],ln[i]);}if( k > n )k = n;mmin = inf;DFS(0,n,0);for(int i=0; i<cnt; i++){memcpy(ltmp,ln,sizeof(ln));for(int j=0; j<k; j++)ltmp[t[i][j]] = ll[t[i][j]];inst_hp_nlogn(ltmp,n,s,len);mmin = min(mmin,area_polygon(s,len));if( mmin == 0 )break;}double ans = area - mmin;printf("%.2lf\n",ans);}return 0;}