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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[N];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;
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