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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】
1 #include<cstdio> 2 #include<cmath> 3 #include<cctype> 4 #include<algorithm> 5 #define ll long long 6 #define maxn 203 7 using namespace std; 8 int n, tot_x, tot_y; 9 ll ans; 10 ll X[maxn], Y[maxn], Renew_X[maxn], Renew_Y[maxn], Map[maxn][maxn]; 11 char buf[1 << 23], *p1 = buf, *p2 = buf, obuf[1 << 23], *O = obuf; 12 bool Mark[maxn][maxn]; 13 struct node { 14 ll x1, x2, y1, y2, xx1, xx2, yy1, yy2; 15 }Pair[maxn]; 16 namespace Ironclad_Programming { 17 #define R register int 18 #define For(i, s, n) for (R i = s; i <= n; ++ i) 19 #define Getch() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1 ++) 20 inline ll read() { 21 int x = 0, f = 1; 22 char ch = Getch(); 23 while(!isdigit(ch)){if (ch == '-')f = -1; ch = Getch();} 24 while(isdigit(ch))x = x * 10 + (ch ^ 48), ch = Getch(); 25 return x * f; 26 } 27 void write(ll x) { 28 if (x > 9) write(x / 10); 29 *O ++ = x % 10 + '0'; 30 } 31 namespace ini { 32 int j = 0; 33 void Discretization() { 34 Renew_X[1] = abs(X[1]); 35 Renew_Y[1] = abs(Y[1]); 36 For (i, 2, 2 * n) { 37 if (X[i] > X[i - 1]) { 38 Renew_X[++ tot_x] = X[i] - X[i - 1]; 39 X[tot_x] = X[i]; 40 } 41 if (Y[i] > Y[i - 1]) { 42 Renew_Y[++ tot_y] = Y[i] - Y[i - 1]; 43 Y[tot_y] = Y[i]; 44 } 45 } 46 } 47 void Convert() { 48 int j; 49 For (i, 1, n) { 50 j = lower_bound(X + 1, X + tot_x, Pair[i].x1) - X; 51 if (X[j] == Pair[i].x1)Pair[i].xx1 = j; 52 j = lower_bound(X + 1, X + tot_x, Pair[i].x2) - X; 53 if (X[j] == Pair[i].x2)Pair[i].xx2 = j; 54 j = lower_bound(Y + 1, Y + tot_y, Pair[i].y1) - Y; 55 if (Y[j] == Pair[i].y1)Pair[i].yy1 = j; 56 j = lower_bound(Y + 1, Y + tot_y, Pair[i].y2) - Y; 57 if (Y[j] == Pair[i].y2)Pair[i].yy2 = j; 58 } 59 } 60 void executive() { 61 n = read(); 62 For (i, 1, n) { 63 Pair[i].x1 = read(), Pair[i].y1 = read(), Pair[i].x2 = read(), Pair[i].y2 = read(); 64 ++ j; 65 X[j] = Pair[i].x1; 66 Y[j] = Pair[i].y1; 67 ++ j; 68 X[j] = Pair[i].x2; 69 Y[j] = Pair[i].y2; 70 } 71 sort(X + 1, X + 2 * n + 1); 72 sort(Y + 1, Y + 2 * n + 1); 73 Discretization(); 74 Convert(); 75 } 76 } 77 void solve() { 78 For (i, 1, tot_y) 79 For (j, 1, tot_x) 80 Map[i][j] = Renew_Y[i] * Renew_X[j]; 81 For (i, 1, n) 82 For (j, Pair[i].yy1 + 1, Pair[i].yy2) 83 For (k, Pair[i].xx1 + 1, Pair[i].xx2) 84 Mark[j][k] = 1; 85 For (i, 1, tot_y) 86 For (j, 1, tot_x) 87 if (Mark[i][j]) 88 ans += Map[i][j]; 89 write(ans); 90 } 91 void Main() { 92 ini::executive(); 93 solve(); 94 fwrite(obuf, O - obuf, 1, stdout); 95 } 96 #undef R 97 #undef For 98 #undef Getch 99 } 100 int main() { 101 Ironclad_Programming::Main(); 102 return 0; 103 }