USACO 5.3 Window Area(模拟)

坐标不全是左上角，右下角的。2Y，怎么可以写的这么麻烦。。。暴力水过。

  1 /*

  2 ID: cuizhe

  3 LANG: C++

  4 TASK: window

  5 */

  6 #include <iostream>

  7 #include <cstdio>

  8 #include <cstring>

  9 #include <cmath>

 10 #include <algorithm>

 11 #include <vector>

 12 #include <string>

 13 #include <queue>

 14 #include <set>

 15 using namespace std;

 16 #define INF 10000000

 17 struct node

 18 {

 19     int x1,y1,x2,y2;

 20 }p[201];

 21 int o[201];

 22 void add(int num,int x1,int y1,int x2,int y2)

 23 {

 24     p[num].x1 = min(x1,x2);

 25     p[num].y1 = max(y1,y2);

 26     p[num].x2 = max(x1,x2);

 27     p[num].y2 = min(y1,y2);

 28 }

 29 int judge(int num,int x1,int y1,int x2,int y2)

 30 {

 31     if(p[num].x1 <= x1&&p[num].y1 >= y1&&x2 <= p[num].x2&&y2 >= p[num].y2)

 32     return 1;

 33     else

 34     return 0;

 35 }

 36 int main()

 37 {

 38     char ch,na;

 39     int i,x1,x2,y1,y2,t,num,bot;

 40     num = 1;bot = 0;

 41     freopen("window.in","r",stdin);

 42     freopen("window.out","w",stdout);

 43     for(i = 0;i <= 200;i ++)

 44     o[i] = -INF;

 45     while(scanf("%c",&ch)!=EOF)

 46     {

 47         if(ch == 'w')

 48         {

 49             scanf("(%c,%d,%d,%d,%d)%*c",&na,&x1,&y1,&x2,&y2);

 50             t = na;

 51             add(t,x1,y1,x2,y2);

 52             o[t] = num ++;

 53         }

 54         else if(ch == 'd')

 55         {

 56             scanf("(%c)%*c",&na);

 57             o[na] = -INF;

 58         }

 59         else if(ch == 't')

 60         {

 61             scanf("(%c)%*c",&na);

 62             o[na] = num ++;

 63         }

 64         else if(ch == 'b')

 65         {

 66             scanf("(%c)%*c",&na);

 67             o[na] = bot--;

 68         }

 69         else if(ch == 's')

 70         {

 71             scanf("(%c)%*c",&na);

 72             set<int> r;

 73             set<int> c;

 74             set<int>::iterator it1;

 75             set<int>::iterator it2;

 76             set<int>::iterator te;

 77             for(i = 0;i <= 130;i ++)

 78             {

 79                 if(o[i] >= o[na])

 80                 {

 81                     r.insert(p[i].x1);

 82                     r.insert(p[i].x2);

 83                     c.insert(p[i].y1);

 84                     c.insert(p[i].y2);

 85                 }

 86             }

 87             int temp1,temp2,sum;

 88             sum = 0;

 89             for(it1 = r.begin();it1 != r.end();it1 ++)

 90             {

 91                 te = ++ it1;

 92                 temp1 = *te;

 93                 it1 --;

 94                 if(te == r.end()) break;

 95 

 96                 for(it2 = c.begin();it2 != c.end();it2 ++)

 97                 {

 98                     te = ++it2;

 99                     temp2 = *te;

100                     it2 --;

101                     if(te == c.end()) break;

102                     if(judge(na,*it1,temp2,temp1,*it2))

103                     {

104                         for(i = 0;i <= 128;i ++)

105                         {

106                             if(o[i] > o[na])

107                             {

108                                 if(judge(i,*it1,temp2,temp1,*it2))

109                                 break;

110                             }

111                         }

112                         if(i != 129)

113                         {

114                             sum += (temp1 - *it1)*(temp2 - *it2);

115                         }

116                     }

117                 }

118             }

119             sum = (p[na].x2-p[na].x1)*(p[na].y1-p[na].y2) - sum;

120             printf("%.3lf\n",sum*100.0/(p[na].x2-p[na].x1)/(p[na].y1-p[na].y2));

121         }

122     }

123     return 0;

124 }