poj 1151 线段树 离散化 扫描线

题目链接：http://poj.org/problem?id=1151

题目大意：在一个平面里有许多矩形，给你每个矩形两对角的坐标，求矩形面积交。

参考：http://blog.csdn.net/new_c_yuer/article/details/6013747

代码：

/**每次得到一个要更新的区间，都单点更新*/#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>using namespace std;#define maxn 205struct Line{      double xl, xr, y;      int tp;      Line(){}      Line(double xl, double xr, double y, int tp):            xl(xl), xr(xr), y(y), tp(tp){}      bool operator < (const Line& b)const      {            return y<b.y;      }}line[maxn];double x[maxn];struct node{      int l, r, tp;      double sum;}v[maxn<<2];void build(int l, int r, int n){      v[n].l = l, v[n].r = r;      v[n].tp = 0;      v[n].sum = 0;      if (l==r)            return ;      int mid = (v[n].l + v[n].r) >> 1;      build(l, mid, n<<1);      build(mid+1, r, n<<1|1);}void update(int n, int xi, int tp){      if (v[n].l==v[n].r)      {            v[n].tp += tp;            if (v[n].tp)                  v[n].sum = x[xi+1] - x[xi];            if (!v[n].tp)                  v[n].sum = 0;            return ;      }      int mid = (v[n].l + v[n].r) >> 1;      if (xi<=mid)            update(n<<1, xi, tp);      else            update(n<<1|1, xi, tp);      v[n].sum = v[n<<1].sum + v[n<<1|1].sum;}int main(){      int n, T = 0;      while (~scanf("%d", &n) && n)      {            int num = 0;            for (int i=0; i<n; ++i)            {                  double x1, x2, y1, y2;                  scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);                  line[++num] = Line(x1, x2, y1, -1);                  x[num] = x1;                  line[++num] = Line(x1, x2, y2, 1);                  x[num] = x2;            }            sort(x+1, x+num+1);            sort(line+1, line+num+1);//            for (int i=1; i<=num; ++i)//                  printf("%.2f ", x[i]);//            printf("\n");            double ans = 0;            build(1, num, 1);            for (int i=1; i<=num; ++i)            {                  int l = lower_bound(x+1, x+num+1, line[i].xl) - x;                  int r = lower_bound(x+1, x+num+1, line[i].xr)- x - 1;                  for (int j=l; j<=r; ++j)                        update(1, j, line[i].tp);                  ans += v[1].sum * (line[i+1].y - line[i].y);            }            printf("Test case #%d\nTotal explored area: %.2f\n\n", ++T, ans);      }      return 0;}

/**线段树的建立与之前的有点不同每次得到一个要更新的区间，成段更新*/#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define maxn 205struct Line{    double xr, xl, y;    int tp;    Line(){}    Line(double xl, double xr, double y, int tp):        xl(xl), xr(xr), y(y), tp(tp){}    bool operator < (const Line& b)const    {        return y<b.y;    }}line[maxn];double x[maxn];struct node{    int l, r, tp;    double sum;}v[maxn<<2];void build(int l, int r, int n){    v[n].l = l, v[n].r = r;    v[n].tp = 0;    v[n].sum = 0;    if (l+1==r)//        return ;    int mid = (l+r) >> 1;    build(l, mid, n<<1);//    build(mid, r, n<<1|1);//}int find(double key, int n){    int l = 1, r = n, mid;    while (l<=r)    {        mid = (l+r) >> 1;        if (x[mid]==key)            return mid;        else if (x[mid]>key)            r = mid-1;        else            l = mid+1;    }}void pushup(int n){    if (v[n].tp)        v[n].sum = x[v[n].r] - x[v[n].l];    else if (v[n].l+1 == v[n].r)        v[n].sum = 0;    else        v[n].sum = v[n<<1].sum + v[n<<1|1].sum;}void update(int n, int l, int r, int tp){    if (l<=v[n].l && v[n].r<=r)    {        v[n].tp += tp;        pushup(n);        return ;    }    int mid = (v[n].l + v[n].r) >> 1;    if (r<=mid)//        update(n<<1, l, r, tp);    else if (l>=mid)//        update(n<<1|1, l, r, tp);    else    {        update(n<<1, l, mid, tp);        update(n<<1|1, mid, r, tp);    }    pushup(n);}int main(){    int n, T = 0;    while (~scanf("%d", &n) && n)    {        int num = 0;        for (int i=0; i<n; i++)        {            double x1, x2, y1, y2;            scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);            line[++num] = Line(x1, x2, y1, 1);            x[num] = x1;            line[++num] = Line(x1, x2, y2, -1);            x[num] = x2;        }        sort(x+1, x+num+1);        sort(line+1, line+num+1);        build(1, num, 1);//        for (int k=1; k<=2*num-1; ++k)//            printf("(%d, %d) tp=%d  sum=%f\n", v[k].l, v[k].r, v[k].tp, v[k].sum);        double ans = 0;        for (int i=1; i<num; i++)        {            int l = find(line[i].xl, num);            int r = find(line[i].xr, num);            update(1, l, r, line[i].tp);//            printf("%.2f\n", v[1].sum);            ans += v[1].sum * (line[i+1].y - line[i].y);        }        printf("Test case #%d\nTotal explored area: %.2f\n\n", ++T, ans);    }    return 0;}