NKOJ 1353 图形面积

时间限制 : 10000 MS   空间限制 : 65536 KB

问题描述

桌面上放了N个矩形，这N个矩形可能有互相覆盖的部分，求它们组成的图形的面积。(矩形的边都与坐标轴平行)

输入格式

输入第一行为一个数N（1≤N≤100），表示矩形的数量。下面N行，每行四个整数，分别表示每个矩形的左下角和右上角的坐标，坐标范围为  到  之间的整数。

输出格式

输出只有一行，一个整数，表示图形的面积。

样例输入

3
1 1 4 3
2 -1 3 2
4 0 5 2

样例输出

10

【标程】

 #include<cstdio>
 #include<cmath>
 #include<cctype>
 #include<algorithm>
 #define ll long long
 #define maxn 203
 using namespace std;
 int n, tot_x, tot_y;
 ll ans;
 ll X[maxn], Y[maxn], Renew_X[maxn], Renew_Y[maxn], Map[maxn][maxn];
 char buf[ << ], *p1 = buf, *p2 = buf, obuf[ << ], *O = obuf;
 bool Mark[maxn][maxn];
 struct node {
     ll x1, x2, y1, y2, xx1, xx2, yy1, yy2;
 }Pair[maxn];
 namespace Ironclad_Programming {
     #define R register int
     #define For(i, s, n) for (R i = s; i <= n; ++ i)
     #define Getch() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1 ++)
     inline ll read() {
         int x = , f = ;
         char ch = Getch();
         while(!isdigit(ch)){if (ch == '-')f = -; ch = Getch();}
         while(isdigit(ch))x = x *  + (ch ^ ), ch = Getch();
         return x * f;
     }
     void write(ll x) {
         if (x > ) write(x / );
         *O ++ = x %  + '';
     }
     namespace ini {
         int j = ;
         void Discretization() {
             Renew_X[] = abs(X[]);
             Renew_Y[] = abs(Y[]);
             For (i, ,  * n) {
                 if (X[i] > X[i - ]) {
                     Renew_X[++ tot_x] = X[i] - X[i - ];
                     X[tot_x] = X[i];
                 }
                 if (Y[i] > Y[i - ]) {
                     Renew_Y[++ tot_y] = Y[i] - Y[i - ];
                     Y[tot_y] = Y[i];
                 }
             }
         }
         void Convert() {
             int j;
             For (i, , n) {
                 j = lower_bound(X + , X + tot_x, Pair[i].x1) - X;
                 if (X[j] == Pair[i].x1)Pair[i].xx1 = j;
                 j = lower_bound(X + , X + tot_x, Pair[i].x2) - X;
                 if (X[j] == Pair[i].x2)Pair[i].xx2 = j;
                 j = lower_bound(Y + , Y + tot_y, Pair[i].y1) - Y;
                 if (Y[j] == Pair[i].y1)Pair[i].yy1 = j;
                 j = lower_bound(Y + , Y + tot_y, Pair[i].y2) - Y;
                 if (Y[j] == Pair[i].y2)Pair[i].yy2 = j;
             }
         }
         void executive() {
             n = read();
             For (i, , n) {
                 Pair[i].x1 = read(), Pair[i].y1 = read(), Pair[i].x2 = read(), Pair[i].y2 = read();
                     ++ j;
                     X[j] = Pair[i].x1;
                     Y[j] = Pair[i].y1;
                     ++ j;
                     X[j] = Pair[i].x2;
                     Y[j] = Pair[i].y2;
                 }
                 sort(X + , X +  * n + );
                 sort(Y + , Y +  * n + );
             Discretization();
             Convert();
         }
     }
     void solve() {
         For (i, , tot_y)
             For (j, , tot_x)
                 Map[i][j] = Renew_Y[i] * Renew_X[j];
         For (i, , n)
             For (j, Pair[i].yy1 + , Pair[i].yy2)
                 For (k, Pair[i].xx1 + , Pair[i].xx2)
                     Mark[j][k] = ;
         For (i, , tot_y)
             For (j, , tot_x)
                 if (Mark[i][j])
                     ans += Map[i][j];
         write(ans);
     }
     void Main() {
         ini::executive();
         solve();
         fwrite(obuf, O - obuf, , stdout);
     }
     #undef R
     #undef For
     #undef Getch
 }
 int main() {
     Ironclad_Programming::Main();
     return ;
 }