标签:style http color os io for ar 问题
题意:给定一些旧机器管辖范围,现在要放一台进新机器,新机器需要空间1*m,问有多少种放法
思路:转化为求面积并的问题,先考虑横放,对于旧机器左边m个位置肯定不能放,还有边界w-m之后的位置肯定不能放,这样就可以把旧机器向左扩大m个位置,并且边界多上一块m*h大小的面积,这样求出这些面积并再用总面积减去即可,求面积的方法利用线段树扫描线,然后对于竖放再做一遍即可,注意特判m=1的情况
代码:
#include <cstdio> #include <cstring> #include <algorithm> using namespace std; #define lson(x) ((x<<1)+1) #define rson(x) ((x<<1)+2) typedef long long ll; const int N = 100005; struct Line { int l, r, h, flag; Line() {} Line(int l, int r, int h, int flag) { this->l = l; this->r = r; this->h = h; this->flag = flag; } } li[N]; struct Point { int x1, y1, x2, y2; } p[N]; bool cmp(Line a, Line b) { return a.h < b.h; } ll w, h; int n, m, tot, hash[N], hn; int find(int x) { int l = 0, r = hn - 1; while (l < r) { int mid = (l + r) / 2; if (hash[mid] < x) l = mid + 1; else r = mid; } return l; } void gethash() { hn = 0; for (int i = 0; i < tot; i++) { if (li[i].flag == 1) continue; hash[hn++] = li[i].l; hash[hn++] = li[i].r; } sort(hash, hash + hn); int tmp = hn; hn = 1; for (int i = 1; i < tmp; i++) if (hash[i] != hash[i - 1]) hash[hn++] = hash[i]; } struct Node { int l, r, addv, len; Node() {} Node(int l, int r, int addv = 0, int len = 0) { this->l = l; this->r = r; this->addv = addv; this->len = len; } } node[N * 4]; void build(int l, int r, int x = 0) { node[x] = Node(l, r); if (l == r) return; int mid = (l + r) / 2; build(l, mid, lson(x)); build(mid + 1, r, rson(x)); } void pushup(int x) { if (node[x].addv) node[x].len = hash[node[x].r + 1] - hash[node[x].l]; else if (node[x].l == node[x].r) node[x].len = 0; else node[x].len = node[lson(x)].len + node[rson(x)].len; } void insert(int l, int r, int v, int x = 0) { if (node[x].l >= l && node[x].r <= r) { node[x].addv += v; pushup(x); return; } int mid = (node[x].l + node[x].r) / 2; if (l <= mid) insert(l, r, v, lson(x)); if (r > mid) insert(l, r, v, rson(x)); pushup(x); } ll solve() { if (n == 0) return h * (w - m + 1); ll ans = 0; sort(li, li + tot, cmp); gethash(); build(0, hn - 2); for (int i = 0; i < tot; i++) { if (i) ans += (ll)node[0].len * (li[i].h - li[i - 1].h); insert(find(li[i].l), find(li[i].r) - 1, li[i].flag); } return h * w - ans; } ll init() { ll ans = 0; tot = 0; for (int i = 0; i < n; i++) { scanf("%d%d%d%d", &p[i].x1, &p[i].y1, &p[i].x2, &p[i].y2); li[tot++] = Line(max(0, p[i].x1 - m), p[i].x2, p[i].y1 - 1, 1); li[tot++] = Line(max(0, p[i].x1 - m), p[i].x2, p[i].y2, -1); } li[tot++] = Line(max(0, (int)w - m + 1), w, 0, 1); li[tot++] = Line(max(0, (int)w - m + 1), w, h, -1); ans += solve(); tot = 0; for (int i = 0; i < n; i++) { li[tot++] = Line(max(0, p[i].y1 - m), p[i].y2, p[i].x1 - 1, 1); li[tot++] = Line(max(0, p[i].y1 - m), p[i].y2, p[i].x2, -1); } li[tot++] = Line(max(0, (int)h - m + 1), h, 0, 1); li[tot++] = Line(max(0, (int)h - m + 1), h, w, -1); swap(w, h); ans += solve(); if (m == 1) ans /= 2; return ans; } int main() { while (~scanf("%lld%lld%d%d", &w, &h, &n, &m)) { printf("%lld\n", init()); } return 0; }
UVA 1492 - Adding New Machine(线段树),布布扣,bubuko.com
UVA 1492 - Adding New Machine(线段树)
标签:style http color os io for ar 问题
原文地址:http://blog.csdn.net/accelerator_/article/details/38638835