hdu 1542（扫描线）

来源：互联网发布：广州星知电子有限公司编辑：程序博客网时间：2024/04/29 21:18

求面积并，感觉这里的线段树就是起二分查找的作用-_-

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#define MAXN 110using namespace std;struct Line {double y1, y2;double x;int flag;};Line line[MAXN * 2];struct Tree {double y1, y2;double x;int flag;};Tree tree[MAXN * 8];double y[MAXN * 2];int dex;bool cmp(Line a, Line b) {return a.x < b.x;}void Build(int l, int r, int i) {if(l == r) {tree[i].y1 = y[dex++];tree[i].y2 = y[dex];tree[i].flag = 0;return ;}int mid = (l + r) >> 1;Build(l, mid, i << 1);Build(mid + 1, r, i << 1 | 1);tree[i].y1 = tree[i << 1].y1;tree[i].y2 = tree[i << 1 | 1].y2;tree[i].flag = 0;}double Update(double x, double y1, double y2, int flag, int l, int r, int i) {if(l == r) {double ans = 0;if(tree[i].flag > 0) ans += (y2 - y1) * (x - tree[i].x);tree[i].x = x;tree[i].flag += flag;return ans;}int mid = (l + r) >> 1;if(y2 <= tree[i << 1].y2) return Update(x, y1, y2, flag, l, mid, i << 1);else if(y1 >= tree[i << 1 | 1].y1) return Update(x, y1, y2, flag, mid + 1, r, i << 1 | 1);else return Update(x, y1, tree[i << 1].y2, flag, l, mid, i << 1) + Update(x, tree[i << 1 | 1].y1, y2, flag, mid + 1, r, i << 1 | 1);}int main() {int n, tt = 1;while(~scanf("%d", &n) && n) {double x1, x2, y1, y2;dex = 0;for(int i = 0; i < n; i++) {scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);y[dex++] = y1, y[dex++] = y2;line[i << 1].y1 = y1, line[i << 1].y2 = y2, line[i << 1].x = x1, line[i << 1].flag = 1;line[i << 1 | 1].y1 = y1, line[i << 1 | 1].y2 = y2, line[i << 1 | 1].x = x2, line[i << 1 | 1].flag = -1;}sort(y, y + dex);dex = 0;Build(1, 2 * n - 1, 1);double ans = 0;sort(line, line + 2 * n, cmp);for(int i = 0; i < 2 * n; i++) {ans += Update(line[i].x, line[i].y1, line[i].y2, line[i].flag, 1, 2 * n - 1, 1);}printf("Test case #%d\n", tt++);printf("Total explored area: %.2lf\n\n", ans);}return 0;}

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