hdoj-1542 Atlantis

考点

  • 扫描线

题解

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#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e2 + 50;
int n, ytot, ltot;
double dy[2 * maxn];

struct node1 {
double x_, y1_, y2_;
int flag_;
bool operator<(node1 &x) { return x_ < x.x_; }
} line[2 * maxn];

struct node2 {
int status_;
double sum_;
} seg[4 * 2 * maxn];
#define lson (rt << 1)
#define rson (rt << 1 | 1)
#define stat(x) seg[x].status_
#define sum(x) seg[x].sum_

void up(int rt, int l, int r) {
sum(rt) = stat(rt) ? dy[r + 1] - dy[l] : (l == r ? 0 : sum(lson) + sum(rson));
}

void update(int rt, int l, int r, int L, int R, int v) {
if (L <= l && r <= R) {
stat(rt) += v, up(rt, l, r);
return;
}
int mid = (l + r) / 2;
if (L <= mid) update(lson, l, mid, L, R, v);
if (R > mid) update(rson, mid + 1, r, L, R, v);
up(rt, l, r);
}

double work() {
double x1, y1, x2, y2;
for (int i = 1; i <= n; ++i) {
cin >> x1 >> y1 >> x2 >> y2;
dy[++ytot] = y1, dy[++ytot] = y2;
line[++ltot] = {x1, y1, y2, 1}, line[++ltot] = {x2, y1, y2, -1};
}
sort(dy + 1, dy + 1 + ytot), sort(line + 1, line + 1 + ltot);
ytot = unique(dy + 1, dy + 1 + ytot) - 1 - dy;
double ans = 0;
for (int i = 1; i < ltot; ++i) {
y1 = lower_bound(dy + 1, dy + 1 + ytot, line[i].y1_) - dy;
y2 = lower_bound(dy + 1, dy + 1 + ytot, line[i].y2_) - dy;
update(1, 1, ytot - 1, y1, y2 - 1, line[i].flag_);
ans += sum(1) * (line[i + 1].x_ - line[i].x_);
}
return ans;
}

int main() {
int cnt = 0;
while (cin >> n && n) {
printf("Test case #%d\n", ++cnt);
printf("Total explored area: %.2lf\n", work());
ytot = ltot = 0;
memset(seg, 0, sizeof(seg));
}
return 0;
}

思路

矩形面积并的裸题,直接上扫描线板子,不再赘述。