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题目大意
给出N个矩形重叠后的图形,要求出重叠后图形的轮廓线长。
题解
参照1999年国家集训队陈宏的论文《数据结构的选择与算法效率——从IOI98试题PICTURE谈起》。
里面说得很清楚,这里不细讲。
就是把PICTURE抽象出一个统计区间个数的模型。简单粗暴地用线段树+离散化上了。
代码
/* TASK:picture LANG:C++ */ #include <cstdio> #include <cstring> #include <algorithm> #include <map> using namespace std; const int MAXN = 5005; const int RANGE = 10001; struct Matrix { int x[2], y[2]; }matrix[MAXN]; struct Line { int h, l, r, delx; bool operator < (const Line &b) const { return h < b.h; } }line[MAXN << 1]; int n, ans, nl, xcnt, ycnt; int leftv[RANGE << 5], rightv[RANGE << 5], lazytag[RANGE << 5], minv[RANGE << 5], maxv[RANGE << 5]; void maintain(int o) { int lc = o << 1, rc = lc + 1; leftv[o] = leftv[lc]; rightv[o] = rightv[rc]; minv[o] = min(minv[lc], minv[rc]); maxv[o] = max(maxv[lc], maxv[rc]); } void pushdown(int o) { int lc = o << 1, rc = lc + 1; if (lazytag[o]) { minv[lc] += lazytag[o]; maxv[lc] += lazytag[o]; minv[rc] += lazytag[o]; maxv[rc] += lazytag[o]; leftv[lc] += lazytag[o]; leftv[rc] += lazytag[o]; rightv[lc] += lazytag[o]; rightv[rc] += lazytag[o]; lazytag[lc] += lazytag[o]; lazytag[rc] += lazytag[o]; lazytag[o] = 0; } } void update(int o, int L, int R, int y1, int y2, int dx) { if (y1 <= L && R <= y2) { leftv[o] += dx; rightv[o] += dx; minv[o] += dx; maxv[o] += dx; lazytag[o] += dx; return; } pushdown(o); int mid = (L + R) >> 1; int lc = o << 1, rc = lc + 1; if (mid >= y1) update(lc, L, mid, y1, y2, dx); if (mid + 1 <= y2) update(rc, mid + 1, R, y1, y2, dx); maintain(o); } int query(int o, int L, int R) { if (minv[o] > 0) return 1; if (maxv[o] == 0) return 0; pushdown(o); int mid = (L + R) >> 1; int lc = o << 1, rc = lc + 1; int a = query(lc, L, mid), b = query(rc, mid + 1, R); if (rightv[lc] && leftv[rc]) return a + b - 1; else return a + b; } void work(int cur) { nl = 0; for (int i = 0; i < n; ++i) if (cur == 0) { line[nl].h = matrix[i].x[0]; line[nl].l = matrix[i].y[0]; line[nl].r = matrix[i].y[1]; line[nl++].delx = 1; line[nl].h = matrix[i].x[1]; line[nl].l = matrix[i].y[0]; line[nl].r = matrix[i].y[1]; line[nl++].delx = -1; } else { line[nl].h = matrix[i].y[0]; line[nl].l = matrix[i].x[0]; line[nl].r = matrix[i].x[1]; line[nl++].delx = 1; line[nl].h = matrix[i].y[1]; line[nl].l = matrix[i].x[0]; line[nl].r = matrix[i].x[1]; line[nl++].delx = -1; } sort(line, line + nl); memset(leftv, 0, sizeof(leftv)); memset(rightv, 0, sizeof(rightv)); memset(maxv, 0, sizeof(maxv)); memset(minv, 0, sizeof(minv)); memset(lazytag, 0, sizeof(lazytag)); memset(minv, 0, sizeof(minv)); for (int i = 0; i < nl - 1; ++i) { update(1, 1, 4 * RANGE, 2 * (line[i].l + RANGE - 1), 2 * (line[i].r + RANGE - 1), line[i].delx); ans += query(1, 1, 4 * RANGE) * 2 * (line[i + 1].h - line[i].h); } } int main() { freopen("picture.in", "r", stdin); freopen("picture.out", "w", stdout); scanf("%d", &n); xcnt = ycnt = 0; for (int i = 0; i < n; ++i) scanf("%d%d%d%d", &matrix[i].x[0], &matrix[i].y[0], &matrix[i].x[1], &matrix[i].y[1]); ans = 0; work(0); work(1); printf("%d\n", ans); return 0; }
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原文地址:http://www.cnblogs.com/albert7xie/p/5325129.html