POJ 2074

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题目链接：http://poj.org/problem?id=2074

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题目大意：求property line上能看到房子全貌的最长段

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题目思路：先求出来每条障碍物线遮挡property line的线段，对线段的左端点进行排序。

从最左一条开始扫描，考察他前面的所有线段中，右端点最远的一条线段和本线段之间的关系，进行分类讨论

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题目细节：

1、障碍线可能在house的上面或者与house平行，或者有可能在property的下面或与其平行。

2、在第一条线段左端点的左边和最右的右端点的右边可能需要计算

3、这题需要用double才能过，且选用c++过不了，需要选g++（木有搞清楚为毛）

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这题改了好几遍，代码的逻辑不清楚了，凑和着看吧。。

源代码：

#include<cstdio>#include<cstring>#include<algorithm>#include<math.h>using namespace std;typedef double T;struct Pt{   T x;   T y;   T len;   Pt(){}   Pt(T px,T py)   {       x = px;       y = py;   }   Pt &operator +(const Pt &p) const   {       Pt c;       c.x = x + p.x;       c.y = y + p.y;       return c;   }   Pt &operator -(const Pt &p) const   {       Pt c;       c.x = x - p.x;       c.y = y - p.y;       return c;   }};struct Line{    Pt a;    Pt b;    int cont;    Line(){}    Line (Pt p1,Pt p2)    {       a = p1;       b = p2;    }};Line line[10000],house,road,cur;bool cmp(Line l1,Line l2){    return l1.a.x < l2.a.x;}Pt seg(Pt a,Pt b){    T x = 0;    x = a.x + (b.x - a.x)*(road.a.y - a.y)/(b.y - a.y);    return Pt(x,road.a.y);}double max(double a,double b){    if(a>b) return a;    else return b;}int main(){    int i = 0,n = 0,j = 0;    T x1,x2,y;    T ans = 0;    T temp = 0;    int flag = 0;    scanf("%lf%lf%lf",&x1,&x2,&y);    while(x1||x2||y)    {        house = Line(Pt(x1,y),Pt(x2,y));        scanf("%lf%lf%lf",&x1,&x2,&y);        road = Line(Pt(x1,y),Pt(x2,y));        if(house.a.y == road.a.y)        {           printf("No View\n");           continue;        }        scanf("%d",&n);        for(i = 0;i<n;i++)        {            scanf("%lf%lf%lf",&x1,&x2,&y);            if(y>=house.a.y || y<=road.a.y)            {               n--;               i--;               continue;            }            cur = Line(Pt(x1,y),Pt(x2,y));            line[i] = Line(seg(house.b,cur.a),seg(house.a,cur.b));        }        sort(line,line+n,cmp);        ans = 0;        if(line[0].a.x>road.a.x && n>0)          ans = line[0].a.x - road.a.x;        for(i = 1;i<n;i++)        {           if(line[i].a.x<=road.a.x) continue;           temp = road.a.x;           for(j = 0;j<i;j++)           {               if(line[j].b.x > temp) temp = line[j].b.x;           }           if(temp>line[i].a.x) continue;           else           {               if(temp>road.b.x) break;               else if(line[i].a.x>road.b.x)                   {                      ans = max(road.b.x - line[i-1].a.x,ans);                      break;                   }                    else ans = max(line[i].a.x-temp,ans);           }        }        if(n == 1)        {            double a = 0,b = 0;            double c = road.b.x - road.a.x;            if(line[0].a.x>road.a.x) a = line[0].a.x - road.a.x;            if(line[0].b.x<road.b.x) b = road.b.x - line[0].b.x;            if(a>c) a = c;            if(b>c) b = c;            ans = max(a,b);        }        else        {            flag = 0;            for(i = 0;i<n;i++)            {               if(line[i].b.x>=road.b.x)               {                   flag = 1;                   break;               }            }            if(!flag) ans = max(road.b.x - line[n-1].b.x,ans);        }        if(ans == 0.0)          printf("No View\n");        else          printf("%.2f\n",ans);        scanf("%lf%lf%lf",&x1,&x2,&y);    }    return 0;}