标签：java struct sts play width tor starting memory for
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23452 Accepted Submission(s): 9278
Problem Description
There are several ancient Greek texts that contain descriptions of the fabled island Atlantis. Some of these texts even include maps of parts of the island. But unfortunately, these maps describe different regions of Atlantis. Your friend Bill has to know the total area for which maps exist. You (unwisely) volunteered to write a program that calculates this quantity.
Input
The input file consists of several test cases. Each test case starts with a line containing a single integer n (1<=n<=100) of available maps. The n following lines describe one map each. Each of these lines contains four numbers x1;y1;x2;y2 (0<=x1<x2<=100000;0<=y1<y2<=100000), not necessarily integers. The values (x1; y1) and (x2;y2) are the coordinates of the top-left resp. bottom-right corner of the mapped area.
The input file is terminated by a line containing a single 0. Don’t process it.
Output
For each test case, your program should output one section. The first line of each section must be “Test case #k”, where k is the number of the test case (starting with 1). The second one must be “Total explored area: a”, where a is the total explored area (i.e. the area of the union of all rectangles in this test case), printed exact to two digits to the right of the decimal point.
Output a blank line after each test case.
Sample Input
2 10 10 20 20 15 15 25 25.5 0
Sample Output
Test case #1
Total explored area: 180.00
分析：模板题
代码：
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int maxn = 110; 4 5 struct edge 6 { 7 double l, r; 8 double h; 9 int pos; 10 bool operator < (const edge &a)const 11 { 12 return h < a.h; 13 } 14 }e[maxn << 1]; 15 16 struct node 17 { 18 int l, r; 19 double sum; 20 int flag; 21 }t[maxn << 3]; 22 int n; 23 double x[maxn << 1]; 24 25 void pushup(int tar) 26 { 27 if (t[tar].flag) t[tar].sum = x[t[tar].r + 1] - x[t[tar].l]; 28 else if (t[tar].l == t[tar].r) t[tar].sum = 0; 29 else t[tar].sum = t[tar << 1].sum + t[tar << 1 | 1].sum; 30 } 31 32 void build(int l, int r, int tar) 33 { 34 t[tar].l = l, t[tar].r = r, t[tar].sum = 0, t[tar].flag = 0; 35 if (l == r) return; 36 int mid = (l + r) >> 1; 37 build(l, mid, tar << 1); 38 build(mid + 1, r, tar << 1 | 1); 39 } 40 41 void update(int l, int r, int v, int tar) 42 { 43 if (l == t[tar].l && r == t[tar].r) 44 { 45 t[tar].flag += v; 46 pushup(tar); 47 return; 48 } 49 int mid = (t[tar].l + t[tar].r) >> 1; 50 if (r <= mid) update(l, r, v, tar << 1); 51 else if (l > mid) update(l, r, v, tar << 1 | 1); 52 else update(l, mid, v, tar << 1), update(mid + 1, r, v, tar << 1 | 1); 53 pushup(tar); 54 } 55 56 int main() 57 { 58 int num, cnt; 59 double x1, y1, x2, y2; 60 int cases = 0; 61 62 while (cin >> n && n) 63 { 64 num = cnt = 0; 65 for (int i = 1; i <= n; i++) 66 { 67 scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2); 68 e[++num].l = x1, e[num].r = x2, e[num].h = y1, e[num].pos = -1; 69 e[++num].l = x1, e[num].r = x2, e[num].h = y2, e[num].pos = 1; 70 x[++cnt] = x1, x[++cnt] = x2; 71 } 72 sort(e + 1, e + 1 + num); 73 sort(x + 1, x + 1 + cnt); 74 int tot = unique(x + 1, x + 1 + cnt) - (x + 1); 75 //cout << tot << endl; 76 build(1, tot, 1); 77 double ans = 0; 78 for (int i = 1; i <= num; i++) 79 { 80 int l = lower_bound(x + 1, x + 1 + tot, e[i].l) - x; 81 int r = lower_bound(x + 1, x + 1 + tot, e[i].r) - x - 1; 82 //cout << l << " " << r << endl; 83 update(l, r, e[i].pos, 1); 84 ans += (e[i + 1].h - e[i].h) * t[1].sum; 85 } 86 printf("Test case #%d\nTotal explored area: %.2f\n\n", ++cases, ans); 87 } 88 }
标签：java struct sts play width tor starting memory for
原文地址：https://www.cnblogs.com/liuwenhan/p/11369048.html
