【扫描线法】&& poj 1177 && hdu 1828
来源:互联网 发布:模拟退火算法 matlab 编辑:程序博客网 时间:2024/06/05 02:37
可以看看这里: http://www.cnblogs.com/Booble/archive/2010/10/10/1847163.html
为了写扫描线, 大概写了有史以来最丑的线段树了。
poj 1177 && hdu 1828 都是求矩形周长并,周长并改一改就可以求面积并了。
其实思想并不复杂,将x维排序,将x维上的2n条线段作为事件,每个事件统计与上个事件之间的所占周长长度。
统计周长长度有点烦, 在y维上要用线段树维护: 共有多长的线段被覆盖,以及共有多少“团”线段。
每次加的答案就是 : 覆盖值得改变量 + (当前线段与前一线段的距离) * 2 * (线段"团"数);
至于线段树,离散之后我维护了六个标记(貌似有神犇只要用三个标记+.+),而且合并的时候写的十分丑陋;
维护了这么六个值: 该区间覆盖最小值,最小值是否在左区间/右区间,最小值出现次数以及团数,以及为了维护最小值而保留的另一个标记。
反正就是很丑,写了5KB,难得有线段树写的这么丑。
# include <cstdio># include <cstdlib># include <cmath># include <cstring>using namespace std;const int maxn = 7000;int len[maxn*2], old[maxn*2],left[maxn], right[maxn], low[maxn],high[maxn];int newlow[maxn], newhigh[maxn], num[maxn*2], id[maxn*2];int bj[maxn*4], tmin[maxn*4], tot[maxn*4], repeat[maxn*4];bool covl[maxn*4], covr[maxn*4];int top, mat, ans, h, st, n, i; void sort(int l, int r){int i = l, j = r, d = num[(l + r) >> 1], dj = id[(l + r) >> 1] % 2,tmp;for (;i <= j;){for (;(num[i] < d) || (num[i] == d && (id[i] % 2) < dj);i++);for (;(num[j] > d) || (num[j] == d && (id[j] % 2) > dj);j--);if (i <= j) tmp = num[i], num[i] = num[j], num[j] = tmp, tmp = id[i], id[i] = id[j], id[j] = tmp, i++, j--;}if (i < r) sort (i, r);if (l < j) sort (l, j);}void prepare_y(){int i; mat = 1, top = 0;for (i = 1; i <= n; i++){num[++top] = low[i], id[top] = i * 2;num[++top] = high[i], id[top] = i * 2 +1; }sort(1, top);for (i = 1; i <= top; i++){if (num[i] != num[i - 1]) mat++;if ((id[i] & 1) == 0 ) newlow[id[i] >> 1] = mat, old[mat] = low[id[i] >> 1];else newhigh[id[i] >> 1] = mat, old[mat] = high[id[i] >> 1]; }for (h = 0, st = 1; st <= mat + 1; st <<= 1, h ++);for (i = 1; i <= st * 2; i++) len[i] = 1; for (i = 1; i <= mat-1; i++) len[st + i] = old[i+1] - old[i];for (i = st-1; i>=1; i--) len[i] = len[i << 1]+ len[(i << 1)+1];memset(id, 0, sizeof(id));memset(num, 0, sizeof(num));}void prepare_x(){top = 0;for (i = 1; i <= n; i++){num[++top] = left[i], id[top] = i * 2;num[++top] = right[i], id[top] = i * 2 +1;}sort(1, top);}void origin(){memset(bj, 0, sizeof(bj));memset(tmin, 0, sizeof(tmin));memset(covl, true, sizeof(covl));memset(covr, true, sizeof(covr));for (int i = 1; i <= st * 2; i++) repeat[i] = 1, tot[i] = 1;for (int i = 1; i <= mat; i++) tot[i +st] = len[i+st];for (int i = st - 1; i >= 1; i--) tot[i] = tot[i << 1] + tot[(i << 1 )+1];}int min(int x, int y){return x < y ? x: y;}void up(int x){for (; x>= 1; x >>=1){tmin[x] = min(tmin[x << 1], tmin[(x << 1) +1]);if (tmin[x << 1] == tmin[(x << 1) +1]){repeat[x] = repeat[x << 1] + repeat[(x << 1) +1] - (covr[x << 1] && covl[(x << 1)+1]);tot[x] = tot[x << 1] + tot[(x << 1 )+1];covl[x] = covl[x << 1], covr[x] = covr[(x << 1) +1]; }else if (tmin[x << 1] < tmin[(x << 1) +1]){repeat[x] = repeat[x << 1];tot[x] = tot[x << 1];covl[x] = covl[x << 1], covr[x] = false;}else{repeat[x] = repeat[(x << 1)+1];tot[x] = tot[(x << 1) +1];covr[x] = covr[(x << 1)+1]; covl[x] = false;}}}void down(int x){ int i, now; for (i = h; i>0; i--) {now = x >> i;if (bj[now] != 0) {bj[now << 1]+= bj[now], bj[(now << 1) +1]+= bj[now];tmin[now << 1]+= bj[now], tmin[(now << 1)+1]+= bj[now];bj[now]= 0; } }}void change(int l, int r, int d){l = l+st-1, r = r+st+1;int ll = l >> 1, rr = r >> 1;down(l); down(r);for (;(l ^ r) != 1; l >>= 1, r >>= 1){if ((l & 1) == 0) bj[l+1]+=d, tmin[l+1]+=d;if ((r & 1) == 1) bj[r-1]+=d, tmin[r-1]+=d;}up(ll); up(rr); }void help(){ for (int i = 1; i <= st; i++) down(i+st);}int main(){freopen("1177.in", "r", stdin);freopen("1177.out", "w", stdout);while (scanf("%d", &n) != EOF){for (i = 1; i <= n; i++) scanf("%d%d%d%d", &left[i], &low[i], &right[i], &high[i]);prepare_y();prepare_x();origin();ans = high[id[1] >> 1] - low[id[1] >> 1];change(newlow[id[1] >> 1], newhigh[id[1] >> 1]-1, 1);for (i = 2; i <= top; i++){ //help();int who = id[i] >> 1;int a1 = len[1]- tot[1] , a2 = repeat[1];if ((who << 1)== id[i]) change(newlow[who], newhigh[who]-1, 1);else change(newlow[who], newhigh[who]-1, -1);int b1 = len[1]- tot[1] ;ans += abs(a1 - b1) + (num[i] - num[i-1]) * 2 * (a2 - 1); }printf("%d\n", ans); }return 0;}
ps: hdu上居然是多组数据,贡献了六七个wa。
附带一个poj1151 求矩形面积并的, 稍微好些一点;
# include <cstdlib># include <cstdio># include <cmath># include <cstring>using namespace std;const int maxn = 200;int id[maxn];double num[maxn], left[maxn], right[maxn], low[maxn], high[maxn], old[maxn*2], tot[maxn*4], len[maxn*4];int tmin[maxn*4], bj[maxn*4], newlow[maxn], newhigh[maxn];int test, h, st, n, top, mat;double ans;void sort(int l, int r){double d = num[(l + r)>> 1], tmpd;int i = l, j = r, tmp, dj = id[(l + r)>> 1] & 1;for (;i <= j;){for (;num[i] < d || (num[i] == d && (id[i] & 1) < dj); i++); for (;num[j] > d || (num[j] == d && (id[j] & 1) > dj); j--);if (i <= j) tmpd=num[i], num[i]=num[j], num[j]=tmpd, tmp=id[i], id[i]=id[j], id[j]=tmp, i++,j--;}if (i < r) sort(i, r);if (l < j) sort(l, j); }void prepare_y(){memset(num, 0, sizeof(num));memset(id, 0, sizeof(id));int i; top = 0; mat = 1;for (i = 1; i <= n; i++){num[++top] = low[i]; id[top] = i * 2;num[++top] = high[i]; id[top] = i * 2 + 1;}sort(1, top);for (i = 1; i <= top; i++){if (num[i] != num[i-1]) mat++;if ((id[i] & 1) == 0) newlow[id[i] >> 1] = mat, old[mat] = low[id[i] >> 1];else newhigh[id[i] >> 1] = mat, old[mat] = high[id[i] >> 1]; }for (st = 1, h = 0; st <= mat +1; st <<= 1, h++);for (i = 1; i <= st*2; i++) len[i] = 1;for (i = 1; i <= mat-1; i++) len[i+st] = old[i+1] - old[i];for (i = st-1; i >= 1; i--) len[i] = len[i << 1] + len[(i << 1)+1];}void prepare_x(){memset(num, 0, sizeof(num));memset(id, 0, sizeof(id));int i; top = 0; mat = 1;for (i = 1; i <= n; i++){num[++top] = left[i], id[top] = i * 2;num[++top] = right[i], id[top] = i * 2 +1;}sort(1, top);}void origin(){int i; ans = 0; memset(tmin, 0, sizeof(tmin));memset(bj, 0 ,sizeof(bj));for (i = 1; i <= st*2; i++) tot[i] = len[i];}void read(){int i; scanf("%d", &n);for (i = 1; i <= n; i++) scanf("%lf%lf%lf%lf", &left[i], &low[i], &right[i], &high[i]);prepare_y();prepare_x();origin();}void down(int x){int i;for (i = h; i > 0; i--){int now = x >> i;if (bj[now] != 0){bj[now << 1]+= bj[now]; bj[(now << 1)+1]+= bj[now];tmin[now << 1]+=bj[now]; tmin[(now << 1)+1]+= bj[now];bj[now] = 0; }}}int min(int x, int y){return x < y ? x:y ;}void up(int x){for (;x > 0; x >>=1){tmin[x] = min(tmin[x << 1], tmin[(x << 1)+1]); if (tmin[x << 1] == tmin[(x << 1)+1]) tot[x] = tot[x << 1] + tot[(x << 1)+1]; else if (tmin[x << 1] < tmin[(x << 1)+1]) tot[x] = tot[x << 1]; else tot[x] = tot[(x << 1)+1];}}void change(int l, int r, int d){l = l+st-1; r= r+st+1; int ll = l >> 1, rr = r >> 1;down(l); down(r);for (;(l ^ r) != 1; l>>=1, r>>=1){if ((l & 1) == 0) tmin[l+1]+= d, bj[l+1]+= d;if ((r & 1) == 1) tmin[r-1]+= d, bj[r-1]+= d;}up(ll); up(rr);}int i; int main(){freopen("1151.in", "r", stdin);freopen("1151.out", "w", stdout);for (test = 1;;test++){read();if ( n == 0) break;change(newlow[id[1] >> 1], newhigh[(id[1]>>1)]-1, 1);for (i = 2; i <= top; i++){ans += (num[i] - num[i-1]) * (len[1] - tot[1]);if ((id[i] & 1) == 0) change(newlow[id[i] >> 1], newhigh[(id[i]>>1)]-1, 1);else change(newlow[id[i] >> 1], newhigh[(id[i]>>1)]-1, -1);}printf("Test case #%d\n", test);printf("Total explored area: %.2lf\n\n", ans);}return 0;}
- 【扫描线法】&& poj 1177 && hdu 1828
- POJ 1177 Picture & hdu 1828 Picture(扫描线)
- poj 1177 / hdu 1828 线段树 离散化 扫描线
- HDU 1828 / POJ 1177 Picture 初涉扫描线
- POJ 1177 & HDU 1828 Picture(扫描线 + 求周长)
- hdu 1828(poj 1177)Picture(线段树+扫描线)(轮廓线)
- HDU 1828 && POJ 1177 Picture(线段树+扫描线+离散化)
- poj 1177 || HDU 1828 Picture (线段树扫描线求 图形并的周长)
- HDU 1828 (扫描线)
- POJ 1151 HDU 1542 Atlantis(扫描线)
- hdu 1828 Picture(扫描线)
- Picture hdu 1828 扫描线
- POJ 1177 Pictures(HDU 1828) (线段树+离散化+线段扫描)
- HDU 1542 && POJ 1151 Atlantis(线段树+扫描线)
- HDU 1542 & POJ 1151 Atlantis【线段树扫描线】
- poj 1151 hdu 1542 atlantis (线段树+扫描线)
- poj 3832&&hdu 3265 扫描线+线段树
- poj 1151 & hdu 1542 Atlantis(线段树,扫描线)
- 统计预报方法
- 寒假-01,最大公因子
- 海思Hi3520开发环境搭建
- Wp7 中 webbrowser 页面跳转
- 在线实时生成Excel文件流供下载 (SSH)
- 【扫描线法】&& poj 1177 && hdu 1828
- git 忽略已经被track的文件
- 网络编程:ASP.JSP.PHP三种技术比较
- .net 学习顺序
- 教师和教务人员的好帮手【教务助手】升级至3.07
- POJ 2449 A* K短路
- mciWnd播放视频音频的操作。
- ubuntu修改swap
- 指针常量和常量指针