http://acm.hdu.edu.cn/showproblem.php?pid=1542 矩形面积的并 线段树 ＋ 扫描线 ＋ 离散化

#include <iostream>#include <cstdio>#include <algorithm>#define LL(x) ((x) << 1)#define RR(x) ((x) << 1 | 1)using namespace std;const int N = 100005;struct Seg_tree{int l, r;int st;double sum;int mid(){return (l + r) >> 1;}}tree[4 * N];struct Line{double high;double l, r;int st;}line[N];double pos[N];int len;inline bool cmp(const Line &a, const Line &b){return a.high < b.high;}inline void Build(int l, int r, int node){tree[node].l = l;tree[node].r = r;tree[node].st = 0;tree[node].sum = 0;if(l == r) return;int mid= (l + r) >> 1;Build(l, mid, LL(node));Build(mid + 1, r, RR(node));}inline void Update(int l, int r, int node, int st){if(l <= tree[node].l && tree[node].r <= r){tree[node].st += st;if(tree[node].st){tree[node].sum = pos[tree[node].r + 1] - pos[tree[node].l];} else if(tree[node].l == tree[node].r){tree[node].sum = 0;} else {tree[node].sum = tree[LL(node)].sum + tree[RR(node)].sum;}return ;}int mid = tree[node].mid();if(r <= mid){Update(l, r, LL(node), st);} else if(l > mid){Update(l, r, RR(node), st);} else{Update(l, mid, LL(node), st);Update(mid + 1, r, RR(node), st);}if(tree[node].st){tree[node].sum = pos[tree[node].r + 1] - pos[tree[node].l];} else if(tree[node].l == tree[node].r){tree[node].sum = 0;} else {tree[node].sum = tree[LL(node)].sum + tree[RR(node)].sum;}}inline int Bin(double x){int low = 1;int high = len - 1;while (low <= high){int mid = (low + high) >> 1;//printf("mid :: %d/n", mid);if(pos[mid] == x) {return mid;} else if (pos[mid] < x){low = mid + 1;} else{high = mid - 1;}}return -1;}int main(){int n;int Case = 1;while(scanf("%d", &n) != EOF && n){double x1, y1, x2, y2;int tt = 1;for(int i = 1; i <= n; i++){scanf("%lf %lf %lf %lf", &x1, &y1, &x2, &y2);//int ll = LL(i);//int rr = RR(i);int ll = tt;int rr = tt + 1;line[ll].high = y1;line[rr].high = y2;line[ll].st = 1;line[rr].st = -1;pos[ll] = line[ll].l = line[rr].l = x1;pos[rr] = line[rr].r = line[ll].r = x2;tt += 2;}//int tt = RR(n);//printf("tt :: %d/n", tt);sort(&pos[1], &pos[1] + tt - 1);sort(&line[1], &line[1] + tt - 1, cmp);len = 1;for(int i = 1; i < tt; i++){//if(pos[i] != pos[i + 1])if(i == 1 || pos[i] != pos[i - 1])pos[len++] = pos[i];}//for(int i = 1; i < tt; i++){//printf("line :: %0.2lf %0.2lf/n", line[i].l, line[i].r);//}//printf("len :: %d/n", len);Build(1 , len, 1);double ans = 0;for(int i = 1; i < tt ; i++){int l = Bin(line[i].l);int r = Bin(line[i].r) - 1;//printf("line :: %0.2lf %0.2lf/n", line[i].l, line[i].r);//printf("i :: %d L :: %d R :: %d/n", i, l, r);//puts("==");Update(l, r, 1, line[i].st);ans += tree[1].sum * (line[i +1].high - line[i].high);//printf("ans :: %0.2lf/n", ans);}printf("Test case #%d/n", Case++);printf("Total explored area: %0.2lf/n/n", ans);}} /* * 　　线段树就矩形面积的并 * 　线段树　＋　扫描线　＋　离散化 * http://acm.pku.edu.cn/JudgeOnline/problem?id=1151 * 　　　　　　　　　　　　　　　　　　　　　　　　　　　 */#include <iostream>#include <cstdio>#include <algorithm>#define LL(x) ((x) << 1)#define RR(x) ((x) << 1 | 1)using namespace std;const int N = 100005;struct Seg_tree {int l, r;int st;//表示区间背覆盖的层数double sum;int mid() {return (l + r) >> 1;}} tree[4 * N];struct Line {double high;double l, r;int st;//　　st = 1 表示下边　ｓｔ　= -1 表示上边} line[N];double pos[N];int len;inline bool cmp(const Line &a, const Line &b) {return a.high < b.high;}inline void Build(int l, int r, int node) {tree[node].l = l;tree[node].r = r;tree[node].st = 0;tree[node].sum = 0;if (l == r)return;int mid = (l + r) >> 1;Build(l, mid, LL(node));Build(mid + 1, r, RR(node));}inline void Update(int l, int r, int node, int st) {if (l <= tree[node].l && tree[node].r <= r) {tree[node].st += st;/* printf("node :: %d l :: %d tree[node].l ::%d tree[noode].r :: %d r :: %d /n",node, l, tree[node].l, tree[node].r, r);printf("tree[node].st :: %d/n", tree[node].st);*/if (tree[node].st) {// 为何要 +１？？tree[node].sum = pos[tree[node].r + 1] - pos[tree[node].l];//puts("1");} else if (tree[node].l == tree[node].r) {tree[node].sum = 0;//puts("2");} else {//printf("tree[node].sum :: %0.2lf/n", tree[node].sum);//printf("tree[LL(node)].sum :: %0.2lf tree[RR(node)].sum :: %0.2lf/n",//tree[LL(node)].sum ,tree[RR(node)].sum);tree[node].sum = tree[LL(node)].sum + tree[RR(node)].sum;//puts("3");}return;}int mid = tree[node].mid();/* if(r <= mid){ Update(l, r, LL(node), st); } else if(l > mid){ Update(l, r, RR(node), st); } else{ Update(l, mid, LL(node), st); Update(mid + 1, r, RR(node), st); } */if (l <= mid)Update(l, r, LL(node), st);if (mid < r)Update(l, r, RR(node), st);if (tree[node].st) { // 为何要 +１？？tree[node].sum = pos[tree[node].r + 1] - pos[tree[node].l];} else if (tree[node].l == tree[node].r) {tree[node].sum = 0;} else {tree[node].sum = tree[LL(node)].sum + tree[RR(node)].sum;}}inline int Bin(double x) {int low = 0;int high = len - 1;while (low <= high) {int mid = (low + high) >> 1;//printf("mid :: %d/n", mid);if (pos[mid] == x) {return mid;}if (pos[mid] < x) {low = mid + 1;} else {high = mid - 1;}}return -1;}int main() {int n;int Case = 1;while (scanf("%d", &n) != EOF && n) {double x1, y1, x2, y2;int tt = 0;for (int i = 0; i < n; i++) {scanf("%lf %lf %lf %lf", &x1, &y1, &x2, &y2);//int ll = LL(i);//int rr = RR(i);line[tt].high = y1;line[tt + 1].high = y2;line[tt].st = 1;line[tt + 1].st = -1;pos[tt] = line[tt].l = line[tt + 1].l = x1;pos[tt + 1] = line[tt].r = line[tt + 1].r = x2;tt += 2;}//int tt = RR(n - 1);sort(pos, pos + tt);sort(line, line + tt, cmp);len = 0;for (int i = 0; i < tt; i++) {//if(pos[i] != pos[i + 1])if (i == 0 || pos[i] != pos[i - 1])pos[len++] = pos[i];}Build(0, len - 1, 1);/*for (int i = 0; i < tt; i++) {printf("line :: %0.2lf %0.2lf/n", line[i].l, line[i].r);}for (int i = 0; i < len; i++) {printf("pos :: %0.2lf/n", pos[i]);}*/double ans = 0;for (int i = 0; i < tt - 1; i++) {int l = Bin(line[i].l);int r = Bin(line[i].r) - 1; // 这里为何要-1？？//printf("line :: %0.2lf %0.2lf/n", line[i].l, line[i].r);//printf("i :: %d L :: %d R :: %d/n", i, l, r);//puts("==");if (l > r) { // 可以不加的ans += tree[1].sum * (line[i + 1].high - line[i].high);continue;}Update(l, r, 1, line[i].st);ans += tree[1].sum * (line[i + 1].high - line[i].high);//printf("tree[1].sum :: %0.2lf/n", tree[1].sum);//printf("ans :: %0.2lf/n", ans);}printf("Test case #%d/n", Case++);printf("Total explored area: %0.2lf/n/n", ans);}}