HDU 1542 Atlantis（线段树求矩形面积并）

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;
}