POJ 1151 Atlantis(离散化+暴力)

题目链接

数据范围较小，直接离散后暴力。等等学习一下线段树的思路。

 1 #include <cstdio>

 2 #include <cstring>

 3 #include <cmath>

 4 #include <algorithm>

 5 using namespace std;

 6 #define N 101

 7 #define eps 1e-6

 8 double xr[N],xc[N],yr[N],yc[N];

 9 double x[3*N],y[3*N];

10 int dblcmp(double x)

11 {

12     if(fabs(x) < eps)

13     return 0;

14     return x > 0 ? 1:-1;

15 }

16 int main()

17 {

18     int n,r,c,i,j,k,cas = 1;

19     double ans;

20     while(scanf("%d",&n)!=EOF)

21     {

22         if(!n) break;

23         ans = 0;

24         for(i = 0; i < n; i ++)

25         {

26             scanf("%lf%lf%lf%lf",&xr[i],&yr[i],&xc[i],&yc[i]);

27             x[2*i] = xr[i];

28             x[2*i+1] = xc[i];

29             y[2*i] = yr[i];

30             y[2*i+1] = yc[i];

31         }

32         r = c = 0;

33         sort(x,x+2*n);

34         sort(y,y+2*n);

35         for(i = 1;i < 2*n;i ++)

36         {

37             if(dblcmp(x[i]-x[r]) == 0)

38             continue;

39             x[++r] = x[i];

40         }

41         for(i = 1;i < 2*n;i ++)

42         {

43             if(dblcmp(y[i]-y[c]) == 0)

44             continue;

45             y[++c] = y[i];

46         }

47         for(i = 0;i < r;i ++)

48         {

49             for(j = 0;j < c;j ++)

50             {

51                 for(k = 0;k < n;k ++)

52                 {

53                     if(dblcmp(x[i]-xr[k])>=0&&dblcmp(y[j]-yr[k])>=0&&dblcmp(x[i+1]-xc[k])<=0&&dblcmp(y[j+1]-yc[k])<=0)

54                     break;

55                 }

56                 if(k != n) ans += (x[i+1]-x[i])*(y[j+1]-y[j]);

57             }

58         }

59         printf("Test case #%d\nTotal explored area: %.2lf\n\n",cas++,ans);

60     }

61     return 0;

62 }

 线段树版本主要是看的以下的博客。

http://www.notonlysuccess.com/index.php/segment-tree-complete/comment-page-3/#comment-4251

http://www.cnblogs.com/kane0526/archive/2013/02/26/2934214.html

 

后一个博客的图很不错。HH解释很好，就是把矩形分为上边和下边，从下到上，扫描。cnt记录区间下边比上边多几条，sum记录存在的长度。一段一段的更新，然后计算即可。

2013-06-20

 

  1 #include <cstdio>

  2 #include <cstring>

  3 #include <string>

  4 #include <algorithm>

  5 using namespace std;

  6 #define maxn 5000

  7 #define lson l , m, rt<<1

  8 #define rson m+1, r,rt<<1|1

  9 double que[maxn];

 10 double sum[maxn];

 11 int cnt[maxn];

 12 struct node

 13 {

 14     double lx,rx,y;

 15     int s;

 16     node(){}

 17     node(double a,double b,double c,int d):lx(a),rx(b),y(c),s(d){}

 18     bool operator < (const node &S) const

 19     {

 20         return y < S.y;

 21     }

 22 }mat[maxn];

 23 int bin(double x,int n)

 24 {

 25     int str,end,mid;

 26     str = 0,end = n;

 27     while(str <= end)

 28     {

 29         mid = (str+end)/2;

 30         if(que[mid] == x)

 31         return mid;

 32         else if(que[mid] > x)

 33         end = mid - 1;

 34         else

 35         str = mid + 1;

 36     }

 37     return mid;

 38 }

 39 void pushup(int rt,int l,int r)

 40 {

 41     if(cnt[rt])

 42     sum[rt] = que[r+1] - que[l];

 43     else if(l == r)

 44     sum[rt] = 0;

 45     else

 46     sum[rt] = sum[rt<<1] + sum[rt<<1|1];

 47 }

 48 void update(int L,int R,int c,int l,int r,int rt)

 49 {

 50     if(L <= l&&r <= R)

 51     {

 52         cnt[rt] += c;

 53         pushup(rt,l,r);

 54         return ;

 55     }

 56     int m = (l+r)>>1;

 57     if(L <= m)

 58     update(L,R,c,lson);

 59     if(R > m)

 60     update(L,R,c,rson);

 61     pushup(rt,l,r);

 62 }

 63 int main()

 64 {

 65     int n,num,i,l,r,cas = 1;

 66     double a,b,c,d;

 67     while(scanf("%d",&n)!=EOF)

 68     {

 69         if(!n) break;

 70         num = 0;

 71         for(i = 0;i < n;i ++)

 72         {

 73             scanf("%lf%lf%lf%lf",&a,&b,&c,&d);

 74             que[num] = a;

 75             mat[num ++] = node(a,c,b,1);

 76             que[num] = c;

 77             mat[num ++] = node(a,c,d,-1);

 78         }

 79         sort(que,que+num);

 80         sort(mat,mat+num);

 81         int k = 1;

 82         for(i = 1;i < num;i ++)

 83         {

 84             if(que[i] != que[i-1])

 85             que[k++] = que[i];

 86         }

 87         double ans = 0;

 88         memset(cnt,0,sizeof(cnt));

 89         memset(sum,0,sizeof(sum));

 90         for(i = 0;i < num-1;i ++)

 91         {

 92             l = bin(mat[i].lx,k-1);

 93             r = bin(mat[i].rx,k-1) - 1;

 94             if(l <= r)

 95             update(l,r,mat[i].s,0,k-1,1);

 96             ans += (mat[i+1].y - mat[i].y)*sum[1];

 97         }

 98         printf("Test case #%d\nTotal explored area: %.2f\n\n",cas++,ans);

 99     }

100     return 0;

101 }