转自:http://blog.csdn.net/lwt36/article/details/48908031
HDU 1542 [POJ 1151] Atlantis (矩形面积并)
题意:
求N<=100个矩形的面积并
分析:
- 离散化: 这些技巧都是老生常谈的了, 不然浮点数怎么建树, 离散化x坐标就可以了
- 扫描线: 首先把矩形按y轴分成两条边, 上边和下边, 对x轴建树, 扫描线可以看成一根平行于x轴的直线.
从y=0开始往上扫, 下边表示要计算面积+1, 上边表示已经扫过了−1, 直到扫到最后一条平行于x轴的边
但是真正在做的时候, 不需要完全模拟这个过程, 一条一条边地插入线段树就好了 - 线段树: 用于动态维护扫描线在往上走时, x轴哪些区域是有合法面积的
- ps:这种线段树是不用lazy的, 因为不用push_down, 为啥不用push_down, 因为没有查询操作
扫描线扫描的过程(建议配合代码模拟)
ps:无论说的再好,都不如自己在纸上模拟一遍扫描的过程,我自己学的时候模拟了很多遍
以下图转载自@kk303的博客
初始状态
扫到最下边的线, 点1→3更新为1
扫到第二根线, 此时S=lcnt!=0∗h两根线之间, 得到绿色的面积, 加到答案中去, 随后更新计数
同上, 将黄色的面积加到答案中去
同上, 将灰色的面积加到答案中去
同上, 将紫色的面积加到答案中去
同上, 将蓝色的面积加到答案中去
#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << " "#define prln(x) cout << #x << " = " << x << endlconst int N = 205, INF = 0x3f3f3f3f, MOD = 1e9 + 7;int n;struct Seg { double l, r, h; int d; Seg() {} Seg(double l, double r, double h, int d): l(l), r(r), h(h), d(d) {} bool operator< (const Seg& rhs) const {return h < rhs.h;}} a[N];int cnt[N << 2]; double sum[N << 2], all[N];#define lson l, m, rt << 1#define rson m + 1, r, rt << 1 | 1void push_up(int l, int r, int rt) { if(cnt[rt]) sum[rt] = all[r + 1] - all[l]; else if(l == r) sum[rt] = 0; else sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];}void update(int L, int R, int v, int l, int r, int rt) { if(L <= l && r <= R) { cnt[rt] += v; push_up(l, r, rt); return; } int m = l + r >> 1; if(L <= m) update(L, R, v, lson); if(R > m) update(L, R, v, rson); push_up(l, r, rt);}int main() {#ifdef LOCAL freopen("in.txt", "r", stdin);#endif ios_base::sync_with_stdio(0); int kase = 0; while(scanf("%d", &n) == 1 && n) { for(int i = 1; i <= n; ++i) { double x1, y1, x2, y2; scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2); a[i] = Seg(x1, x2, y1, 1); a[i + n] = Seg(x1, x2, y2, -1); all[i] = x1; all[i + n] = x2; } n <<= 1; sort(a + 1, a + 1 + n); sort(all + 1, all + 1 + n); int m = unique(all + 1, all + 1 + n) - all - 1; memset(cnt, 0, sizeof cnt); memset(sum, 0, sizeof sum); double ans = 0; for(int i = 1; i < n; ++i) { int l = lower_bound(all + 1, all + 1 + m, a[i].l) - all; int r = lower_bound(all + 1, all + 1 + m, a[i].r) - all; if(l < r) update(l, r - 1, a[i].d, 1, m, 1); ans += sum[1] * (a[i + 1].h - a[i].h); } printf("Test case #%d\nTotal explored area: %.2f\n\n", ++kase, ans); } return 0;}
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HDU 1255 覆盖的面积 (矩形面积交)
题意:
求N<=1000个矩形覆盖至少两次区域的面积,也就是矩形面积交
分析
- 前面的与矩形面积并类似, 不同的是push_up的时候要考虑至少覆盖一次one和至少覆盖两次two的更新
尤其是当前被覆盖了一次的时候, 由于没有push_down操作, 父亲节点的信息是没有同步到儿子节点的, 这样的话push_up就要考虑了. - 父亲被记录覆盖了一次, 但是如果儿子被覆盖过, 这些操作都是在这个父亲这个大区间上的, 就相当于父亲区间被覆盖了至少两次, 所以two和one都要更新
代码
#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << " "#define prln(x) cout << #x << " = " << x << endlconst int N = 2e3 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;int n;struct Seg { double l, r, h; int d; Seg() {} Seg(double l, double r, double h, double d): l(l), r(r), h(h), d(d) {} bool operator< (const Seg& rhs) const { return h < rhs.h; }} a[N];int cnt[N << 2];double one[N << 2], two[N << 2], all[N];#define lson l, m, rt << 1#define rson m + 1, r, rt << 1 | 1void push_up(int l, int r, int rt) { if(cnt[rt] >= 2) two[rt] = one[rt] = all[r + 1] - all[l]; else if(cnt[rt] == 1) { one[rt] = all[r + 1] - all[l]; if(l == r) two[rt] = 0; else two[rt] = one[rt << 1] + one[rt << 1 | 1]; } else { if(l == r) one[rt] = two[rt] = 0; else { one[rt] = one[rt << 1] + one[rt << 1 | 1]; two[rt] = two[rt << 1] + two[rt << 1 | 1]; } }}void update(int L, int R, int v, int l, int r, int rt) { if(L <= l && r <= R) { cnt[rt] += v; push_up(l, r, rt); return; } int m = l + r >> 1; if(L <= m) update(L, R, v, lson); if(R > m) update(L, R, v, rson); push_up(l, r, rt);}int main() {#ifdef LOCAL freopen("in.txt", "r", stdin);#endif ios_base::sync_with_stdio(0); int t; scanf("%d", &t); while(t--) { scanf("%d", &n); for(int i = 1; i <= n; ++i) { double x1, y1, x2, y2; scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2); a[i] = Seg(x1, x2, y1, 1); a[i + n] = Seg(x1, x2, y2, -1); all[i] = x1; all[i + n] = x2; } n <<= 1; sort(a + 1, a + 1 + n); sort(all + 1, all + 1 + n); int m = unique(all + 1, all + 1 + n) - all - 1; memset(cnt, 0, sizeof cnt); memset(one, 0, sizeof one); memset(two, 0, sizeof two); double ans = 0; for(int i = 1; i < n; ++i) { int l = lower_bound(all + 1, all + 1 + m, a[i].l) - all; int r = lower_bound(all + 1, all + 1 + m, a[i].r) - all; if(l < r) update(l, r - 1, a[i].d, 1, m, 1); ans += two[1] * (a[i + 1].h - a[i].h); } printf("%.2f\n", ans); } return 0;}
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HDU 1828 [POJ 1177] Picture(矩形周长并)
题意:
求N<=5000个矩形的轮廓长度,也就是矩形周长并
分析一:
可以用类似矩形面积并的办法, 不过这次我们不乘高, 不算面积罢了.
需要注意的是, 由于周长的线会被重复覆盖, 我们每次需要和上一次的作差.
但是这样仅仅是x轴的, 不过我可以再y轴做一次加起来就可以了
演示x轴求长度和的部分
代码一:
#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << " "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e4 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;int n, m[2];int sum[N << 2], cnt[N << 2], all[2][N];struct Seg { int l, r, h, d; Seg() {} Seg(int l, int r, int h, int d): l(l), r(r), h(h), d(d) {} bool operator< (const Seg& rhs) const {return h < rhs.h;}} a[2][N];#define lson l, m, rt << 1#define rson m + 1, r, rt << 1 | 1void push_up(int p, int l, int r, int rt) { if(cnt[rt]) sum[rt] = all[p][r + 1] - all[p][l]; else if(l == r) sum[rt] = 0; else sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];}void update(int p, int L, int R, int v, int l, int r, int rt) { if(L <= l && r <= R) { cnt[rt] += v; push_up(p, l, r, rt); return; } int m = l + r >> 1; if(L <= m) update(p, L, R, v, lson); if(R > m) update(p, L, R, v, rson); push_up(p, l, r, rt);}int main() {#ifdef LOCAL freopen("in.txt", "r", stdin);#endif ios_base::sync_with_stdio(0); while(scanf("%d", &n) == 1) { for(int i = 1; i <= n; ++i) { int x1, y1, x2, y2; scanf("%d%d%d%d", &x1, &y1, &x2, &y2); all[0][i] = x1, all[0][i + n] = x2; all[1][i] = y1, all[1][i + n] = y2; a[0][i] = Seg(x1, x2, y1, 1); a[0][i + n] = Seg(x1, x2, y2, -1); a[1][i] = Seg(y1, y2, x1, 1); a[1][i + n] = Seg(y1, y2, x2, -1); } n <<= 1; sort(all[0] + 1, all[0] + 1 + n); m[0] = unique(all[0] + 1, all[0] + 1 + n) - all[0] - 1; sort(all[1] + 1, all[1] + 1 + n); m[1] = unique(all[1] + 1, all[1] + 1 + n) - all[1] - 1; sort(a[0] + 1, a[0] + 1 + n); sort(a[1] + 1, a[1] + 1 + n); int ans = 0; for(int i = 0; i < 2; ++i) { int t = 0, last = 0; memset(cnt, 0, sizeof cnt); memset(sum, 0, sizeof sum); for(int j = 1; j <= n; ++j) { int l = lower_bound(all[i] + 1, all[i] + 1 + m[i], a[i][j].l) - all[i]; int r = lower_bound(all[i] + 1, all[i] + 1 + m[i], a[i][j].r) - all[i]; if(l < r) update(i, l, r - 1, a[i][j].d, 1, m[i], 1); t += abs(sum[1] - last); last = sum[1]; } ans += t; } printf("%d\n", ans); } return 0;}
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分析二:
当然我们也可只对x轴做一次扫描线, 只要同时维护y轴竖线(就是求矩形面积并的时候的高)的个数, vtl记录竖线的个数
需要的注意的是竖线重合的情况, 需要再开变量lbd,rbd来判断重合, 避免重复计算
代码二:
#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << " "#define prln(x) cout << #x << " = " << x << endlconst int N = 2e4 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;int n;int sum[N << 2], cnt[N << 2], vtl[N << 2];bool lbd[N << 2], rbd[N << 2];struct Seg { int l, r, h, d; Seg() {} Seg(int l, int r, int h, int d): l(l), r(r), h(h), d(d) {} bool operator< (const Seg& rhs) const {return h < rhs.h;}} a[N];#define lson l, m, rt << 1#define rson m + 1, r, rt << 1 | 1void push_up(int l, int r, int rt) { if(cnt[rt]) { lbd[rt] = rbd[rt] = true; sum[rt] = r + 1 - l; vtl[rt] = 2; } else if(l == r) sum[rt] = vtl[rt] = lbd[rt] = rbd[rt] = 0; else { lbd[rt] = lbd[rt << 1]; rbd[rt] = rbd[rt << 1 | 1]; sum[rt] = sum[rt << 1] + sum[rt << 1 | 1]; vtl[rt] = vtl[rt << 1] + vtl[rt << 1 | 1]; if(rbd[rt << 1] && lbd[rt << 1 | 1]) vtl[rt] -= 2; }}void update(int L, int R, int v, int l, int r, int rt) { if(L <= l && r <= R) { cnt[rt] += v; push_up(l, r, rt); return; } int m = l + r >> 1; if(L <= m) update(L, R, v, lson); if(R > m) update(L, R, v, rson); push_up(l, r, rt);}int main() {#ifdef LOCAL freopen("in.txt", "r", stdin);#endif ios_base::sync_with_stdio(0); while(scanf("%d", &n) == 1) { int Min = 1e4, Max = -1e4; for(int i = 1; i <= n; ++i) { int x1, y1, x2, y2; scanf("%d%d%d%d", &x1, &y1, &x2, &y2); Min = min(Min, x1); Max = max(Max, x2); a[i] = Seg(x1, x2, y1, 1); a[i + n] = Seg(x1, x2, y2, -1); } n <<= 1; sort(a + 1, a + 1 + n); int ans = 0, last = 0; for(int i = 1; i <= n; ++i) { if(a[i].l < a[i].r) update(a[i].l, a[i].r - 1, a[i].d, Min, Max - 1, 1); ans += vtl[1] * (a[i + 1].h - a[i].h); ans += abs(sum[1] - last); last = sum[1]; } printf("%d\n", ans); } return 0;}