标签:abs ace efi archive 次数 bsp int 不重复 com
参考 https://www.cnblogs.com/null00/archive/2012/04/22/2464876.html
#include <stdio.h> #include <algorithm> #define LEN 10000 using namespace std; struct Node { int left; int right; int count;//被覆盖次数 //所包含的区间数量,如三条[1,2],[2,3],[4,5]线段被覆盖,则line=2,因为 [1,2],[2,3]是连续的。 int line;//所包含的区间数量 int lbd;//左端点是否被覆盖 用来辅助对line的计算 int rbd;//右端点是否被覆盖 int m;//测度 ,即覆盖的区间长度,如[2,8]就为6 }node[LEN*4];; struct ScanLine { int x; int y1; int y2; int flag; }scan[LEN];; int y[LEN]; void build(int l, int r, int i) { node[i].left = l; node[i].right = r; node[i].count = 0; node[i].m = 0; node[i].line = 0; if (r - l > 1) { int middle = (l + r)/2; build(l, middle, 2*i + 1); build(middle, r, 2*i + 2); } } //更新测度m void update_m(int i) { if (node[i].count > 0) node[i].m = y[node[i].right] - y[node[i].left]; else if (node[i].right - node[i].left == 1) node[i].m = 0; else { node[i].m = node[2*i + 1].m + node[2*i + 2].m; } } //更新line void update_line(int i) { if (node[i].count > 0) { node[i].lbd = 1; node[i].rbd = 1; node[i].line = 1; } else if (node[i].right - node[i].left == 1) { node[i].lbd = 0; node[i].rbd = 0; node[i].line = 0; } else { node[i].lbd = node[2*i + 1].lbd; node[i].rbd = node[2*i + 2].rbd; node[i].line = node[2*i + 1].line + node[2*i + 2].line - node[2*i + 1].rbd*node[2*i + 2].lbd; } } void insert(int l, int r, int i) { //在这里要取离散化之前的原值进行比较 if (y[node[i].left] >= l && y[node[i].right] <= r) (node[i].count)++; else if (node[i].right - node[i].left == 1) return; else { int middle = (node[i].left + node[i].right)/2; if (r <= y[middle]) insert(l, r, 2*i + 1); else if (l >= y[middle]) insert(l, r, 2*i + 2); else { insert(l, y[middle], 2*i + 1); insert(y[middle], r, 2*i + 2); } } update_m(i); update_line(i); } void remove(int l, int r, int i) { ////在这里要取离散化之前的原值进行比较 if (y[node[i].left] >= l && y[node[i].right] <= r) (node[i].count)--; else if (node[i].right - node[i].left == 1) return; else { int middle = (node[i].left + node[i].right)/2; if (r <= y[middle]) remove(l, r, 2*i + 1); else if (l >= y[middle]) remove(l, r, 2*i + 2); else { remove(l, y[middle], 2*i + 1); remove(y[middle], r, 2*i + 2); } } update_m(i); update_line(i); } bool cmp(ScanLine line1,ScanLine line2) { if (line1.x == line2.x) return line1.flag > line2.flag; return (line1.x < line2.x); } int main() { int n; scanf("%d", &n); int x1, y1, x2, y2; int i = 0; while (n--) { scanf("%d %d %d %d", &x1, &y1, &x2, &y2); scan[i].x = x1; scan[i].y1 = y1; scan[i].y2 = y2; scan[i].flag = 1; y[i++] = y1; scan[i].x = x2; scan[i].y1 = y1; scan[i].y2 = y2; scan[i].flag = 0; y[i++] = y2; } sort(y, y + i); sort(scan, scan + i, cmp); //y数组中不重复的个数 int unique_count = unique(y, y + i) - y; //离散化,建立线段树 build(0, unique_count - 1, 0); int perimeter = 0; int now_m = 0; int now_line = 0; for (int j = 0; j < i; j++) { if (scan[j].flag) insert(scan[j].y1, scan[j].y2, 0); else remove(scan[j].y1, scan[j].y2, 0); if (j >= 1) perimeter += 2*now_line*(scan[j].x - scan[j-1].x); //要减去,因为一个边只能算一次,要减去上一次已经算过的边 perimeter += abs(node[0].m - now_m); now_m = node[0].m; now_line = node[0].line; } printf("%d\n", perimeter); return 0; }
Picture POJ - 1177 线段树+离散化+扫描线 求交叉图像周长
标签:abs ace efi archive 次数 bsp int 不重复 com
原文地址:https://www.cnblogs.com/QingyuYYYYY/p/12296636.html