POJ 1151
题目大意：给出n个矩形（通过给出对角坐标确定），这些矩形间可能有重叠，求除去重叠后覆盖的面积是多少。
解：矩形切割，只是用线段树去维护线段达到logn的效率。
View Code
1 //poj 1151
2 const
3 maxn=1111;
4 inf='1.txt';
5 type
6 type_node=record
7 mid, l, r, flag: longint;
8 x1, x2, cnt: double;
9 end;
10 type_line=record
11 x1, x2, y: double;
12 op: longint;
13 end;
14 var
15 tree: array[0..maxn*4*2]of type_node;
16 x: array[0..maxn*2]of double;
17 line: array[0..maxn*2]of type_line;
18 tot, test, n: longint;
19 ans: double;
20 procedure _add(x1, x2, yy: double; zz: longint);
21 begin
22 inc(tot);
23 line[tot].x1 := x1; line[tot].x2 := x2;
24 line[tot].y := yy; line[tot].op := zz;
25 end;
26
27 procedure _sortline(b, e: longint);
28 var
29 i, j: longint;
30 x: double;
31 k: type_line;
32 begin
33 i := b; j := e; x := line[(i+j)>>1].y;
34 repeat
35 while line[i].y>x do inc(i);
36 while line[j].y<x do dec(j);
37 if i<=j then begin
38 k := line[i]; line[i] := line[j]; line[j] := k;
39 inc(i); deC(j);
40 end;
41 until i>j;
42 if j>b then _sortline(b, j);
43 if i<e then _sortline(i, e);
44 end;
45
46 procedure _sortx(b, e: longint);
47 var
48 i, j: longint;
49 xx, k: double;
50 begin
51 i := b; j := e; xx := x[(i+j)>>1];
52 repeat
53 while x[i]<xx do inc(i);
54 while x[j]>xx do dec(j);
55 if i<=j then begin
56 k := x[i]; x[i] := x[j]; x[j] := k;
57 inc(i); dec(j);
58 end;
59 until i>j;
60 if j>b then _sortx(b, j);
61 if i<e then _sortx(i, e);
62 end;
63
64 procedure _build(ll, rr, kk: longint);
65 begin
66 with tree[kk] do begin
67 flag := 0; cnt := 0;
68 x1 := x[ll]; x2 := x[rr];
69 if ll+1>=rr then exit;
70 mid := (ll+rr)>>1;
71 l := kk << 1; _build(ll, mid, l);
72 r := l + 1; _build(mid, rr, r);
73 end;
74 end;
75
76 procedure _init;
77 var
78 i: longint;
79 x1, x2, y1, y2: double;
80 begin
81 tot := 0; ans := 0;
82 for i := 1 to n do begin
83 readln(x1, y1, x2, y2);
84 _add(x1, x2, y1, -1); x[tot] := x1;
85 _add(x1, x2, y2, 1); x[tot] := x2;
86 end;
87 _sortline(1, tot);
88 _sortx(1, tot);
89 _build(1, tot, 1);
90 end;
91
92 procedure calc(k: longint);
93 begin
94 with tree[k] do begin
95 if flag>0 then begin
96 cnt := x2-x1;
97 exit;
98 end;
99 if (l+r<>0) then cnt := tree[l].cnt+tree[r].cnt;
100 end;
101 end;
102
103 procedure _cover(xx, yy: double; zz, k: longint);
104 begin
105 tree[k].cnt := 0;
106 if (xx<=tree[k].x1)and(tree[k].x2<=yy) then begin
107 inc(tree[k].flag, zz);
108 calc(k);
109 exit;
110 end;
111 if tree[k].mid=0 then exit;
112 if yy<=x[tree[k].mid] then _cover(xx, yy, zz, tree[k].l)
113 else if xx>=x[tree[k].mid] then _cover(xx, yy, zz, tree[k].r)
114 else begin
115 _cover(xx, x[tree[k].mid], zz, tree[k].l);
116 _cover(x[tree[k].mid], yy, zz, tree[k].r);
117 end;
118 calc(k);
119 end;
120
121 procedure _main;
122 var
123 i: longint;
124 begin
125 _cover(line[1].x1, line[1].x2, line[1].op, 1);
126 for i := 2 to tot do begin
127 ans := ans + tree[1].cnt*(line[i-1].y-line[i].y);
128 _cover(line[i].x1, line[i].x2, line[i].op, 1);
129 end;
130 end;
131
132 procedure _print;
133 begin
134 writeln('Test case #', test);
135 writeln('Total explored area: ', ans:0:2);
136 end;
137
138 begin
139 assign(input,inf); reset(input);
140 fillchar(tree, sizeof(tree), 0);
141 test := 0;
142 readln(n);
143 while (n<>0) do begin
144 inc(test);
145 _init;
146 _main;
147 _print;
148 writeln;
149 readln(n);
150 end;
151 end.