HDU 1542 Atlantis（线段树求矩形面积并）

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题意：给出n个矩形的两个端点，求出它们的面积并。

思路：求矩形的面积并模板题。

#include<cstdio>#include<cstring>#include<cmath>#include<cstdlib>#include<iostream>#include<algorithm>#include<vector>#include<map>#include<queue>#include<stack>#include<string>#include<map>#include<set>#include<ctime>#define eps 1e-6#define LL long long#define pii pair<int, int>//#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;const int MAXN = 210;//const int INF = 0x3f3f3f3f;struct Node {int l, r, c;double lf, rf, cnt;} segTree[MAXN*4];struct Line {int c;double x, y1, y2;bool operator < (const Line& E) const {return x < E.x;}} line[MAXN];double y[MAXN];void build(int o, int L, int R) {Node& t = segTree[o];t.l = L;t.r = R;t.c = 0;t.lf = y[L];t.rf = y[R];t.cnt = 0;if(L+1 == R) return;int M =  (L+R) >> 1;build(o<<1, L, M);build((o<<1)|1, M, R);}void cal(int o) {Node& t = segTree[o];if(t.c > 0) {t.cnt = t.rf - t.lf;return;}t.cnt = segTree[o<<1].cnt + segTree[(o<<1)|1].cnt;}void update(int o, Line e) {Node& t = segTree[o];if(e.y1==t.lf && e.y2==t.rf) {t.c += e.c;cal(o);return;}if(e.y2 <= segTree[o<<1].rf) update(o<<1, e);else if(e.y1 >= segTree[(o<<1)|1].lf) update((o<<1)|1, e);else {Line tmp = e;tmp.y2 = segTree[o<<1].rf;update(o<<1, tmp);e.y1 = segTree[(o<<1)|1].lf;update((o<<1)|1, e);}cal(o);}int n, tot;int main() {    //freopen("input.txt", "r", stdin);    int kase = 0;while(cin >> n && n) {tot = 0;for(int i = 1; i <= n; i++) {double x1,x2,y1,y2;scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);++tot;line[tot].y1 = y1;line[tot].y2 = y2;line[tot].x = x1;line[tot].c = 1;y[tot] = y1;++tot;line[tot].y1 = y1;line[tot].y2 = y2;line[tot].x = x2;line[tot].c = -1;y[tot] = y2;}sort(y+1, y+tot+1);sort(line+1, line+tot+1);build(1, 1, tot);double ans = 0;update(1, line[1]);for(int i = 2; i <= tot; i++) {ans += (line[i].x-line[i-1].x) * segTree[1].cnt;update(1, line[i]);}printf("Test case #%d\n", ++kase);printf("Total explored area: %.2f\n\n", ans);}    return 0;}

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