HDU 3340 Rain in ACStar（线段树+几何）

HDU 3340 Rain in ACStar

题目链接

题意：给定几个多边形（3-5边形），然后中间有一些询问，询问一个区间的总面积

思路：多边形分割为梯形，梯形的面积为上底d1 + 下底d2 乘上 高度 / 2，两个梯形面积累加的话，可以等价为上底下底累加，所以就可以用线段树搞了，然后给定的多边形点是按顺序的，可以利用容斥去方便把一个询问拆分成几个询问

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;

const int N = 133333;
int t, n;

struct Point {
int x, y, id;
Point() {}
Point(int x, int y) {
this->x = x;
this->y = y;
this->id = id;
}
void read() {
scanf("%d%d", &x, &y);
}
} p[6];

bool cmpp(Point a, Point b) {
return a.x < b.x;
}

struct OP {
int tp, l, r;
double a, b;
OP() {}
OP(int l, int r, int tp, double a = 0.0, double b = 0.0) {
this->l = l; this->r = r; this->tp = tp;
this->a = a; this->b = b;
}
} op
;

int hash[N * 2], hn, opn;

int find(int x) {
return lower_bound(hash, hash + hn, x) - hash;
}

void build_p(int m) {
p[m++] = p[0];
for (int i = 0; i < m - 1; i++) {
if (p[i].x < p[i + 1].x) op[opn++] = OP(p[i].x, p[i + 1].x, 1, -p[i].y, -p[i + 1].y);
else op[opn++] = OP(p[i + 1].x, p[i].x, 1, p[i + 1].y, p[i].y);
}
}

#define lson(x) ((x<<1)+1)
#define rson(x) ((x<<1)+2)

struct Node {
int l, r;
double d1, d2, s;
void gao(double a, double b) {
d1 += a;
d2 += b;
s += (a + b) * (hash[r + 1] - hash[l]) / 2;
}
} node[N * 8];

void build(int l, int r, int x = 0) {
node[x].l = l; node[x].r = r;
node[x].d1 = node[x].d2 = node[x].s = 0;
if (l == r) return;
int mid = (l + r) / 2;
build(l, mid, lson(x));
build(mid + 1, r, rson(x));
}

double cal(double h1, double h2, double d1, double d2) {
return (d2 * h1 + d1 * h2) / (h1 + h2);
}

void pushdown(int x) {
double h1 = hash[node[lson(x)].r + 1] - hash[node[lson(x)].l];
double h2 = hash[node[rson(x)].r + 1] - hash[node[rson(x)].l];
double d = cal(h1, h2, node[x].d1, node[x].d2);
node[lson(x)].gao(node[x].d1, d);
node[rson(x)].gao(d, node[x].d2);
node[x].d1 = node[x].d2 = 0;
}

void pushup(int x) {
node[x].s = node[lson(x)].s + node[rson(x)].s;
}

void add(int l, int r, double a, double b, int x = 0) {
if (node[x].l >= l && node[x].r <= r) {
double d1 = cal(hash[node[x].l] - hash[l], hash[r + 1] - hash[node[x].l], a, b);
double d2 = cal(hash[node[x].r + 1] - hash[l], hash[r + 1] - hash[node[x].r + 1], a, b);
node[x].gao(d1, d2);
return;
}
int mid = (node[x].l + node[x].r) / 2;
pushdown(x);
if (l <= mid) add(l, r, a, b, lson(x));
if (r > mid) add(l, r, a, b, rson(x));
pushup(x);
}

double query(int l, int r, int x = 0) {
if (node[x].l >= l && node[x].r <= r) return node[x].s;
int mid = (node[x].l + node[x].r) / 2;
double ans = 0;
pushdown(x);
if (l <= mid) ans += query(l, r, lson(x));
if (r > mid) ans += query(l, r, rson(x));
pushup(x);
return ans;
}

int main() {
scanf("%d", &t);
while (t--) {
scanf("%d", &n);
char ss[15];
int l, r, m;
hn = opn = 0;
while (n--) {
scanf("%s", ss);
if (ss[0] == 'Q') {
scanf("%d%d", &l, &r);
hash[hn++] = l; hash[hn++] = r;
op[opn++] = OP(l, r, 0);
}
else {
scanf("%d", &m);
for (int i = 0; i < m; i++) {
p[i].read();
hash[hn++] = p[i].x;
}
build_p(m);
}
}
int tmp = 1;
sort(hash, hash + hn);
for (int i = 1; i < hn; i++)
if (hash[i] != hash[i - 1])
hash[tmp++] = hash[i];
hn = tmp;
build(0, hn - 2);
for (int i = 0; i < opn; i++) {
if (op[i].tp) add(find(op[i].l), find(op[i].r) - 1, op[i].a, op[i].b);
else printf("%.3lf\n", query(find(op[i].l), find(op[i].r) - 1));
}
}
return 0;
}