hdu1542 Atlantis (线段树+矩阵面积并+离散化)
InputThe input file consists of several test cases. Each test case starts with a line containing a single integer n (1<=n<=100) of available maps. The n following lines describe one map each. Each of these lines contains four numbers x1;y1;x2;y2 (0<=x1<x2<=100000;0<=y1<y2<=100000), not necessarily integers. The values (x1; y1) and (x2;y2) are the coordinates of the top-left resp. bottom-right corner of the mapped area.
The input file is terminated by a line containing a single 0. Don’t process it.OutputFor each test case, your program should output one section. The first line of each section must be “Test case #k”, where k is the number of the test case (starting with 1). The second one must be “Total explored area: a”, where a is the total explored area (i.e. the area of the union of all rectangles in this test case), printed exact to two digits to the right of the decimal point.
Output a blank line after each test case.
Sample Input
2
10 10 20 20
15 15 25 25.5
0
Sample Output
Test case #1
Total explored area: 180.00 思路:
扫描线+矩阵面积并,被离散化难到了,之前一直没想到离散化要弄成一个左开右闭的区间 实现代码:
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define mid int m = (l + r) >> 1
const int M = 1e5+;
struct seg{
double l,r,h;
int s;
seg(){}
seg(double a,double b,double c,int d):l(a),r(b),h(c),s(d){}
bool operator < (const seg &cmp) const {
return h < cmp.h;
}
}t[M];
double sum[M<<],x[M<<];
int cnt[M<<];
void pushup(int l,int r,int rt){
if(cnt[rt]) sum[rt] = x[r+] - x[l];
else if(l == r) sum[rt] = ;
else sum[rt] = sum[rt<<] + sum[rt<<|];
} void update(int L,int R,int c,int l,int r,int rt){
if(L <= l&&R >= r){
cnt[rt] += c;
pushup(l,r,rt);
return ;
}
mid;
if(L <= m) update(L,R,c,lson);
if(R > m) update(L,R,c,rson);
pushup(l,r,rt);
} int bin(double key,int n,double x[]){
int l = ;int r = n-;
while(l <= r){
mid;
if(x[m] == key) return m;
else if(x[m] < key) l = m+;
else r = m-;
}
return -;
}
int main()
{
int n,cas = ;
double a,b,c,d;
while(~scanf("%d",&n)&&n){
int m = ;
while(n--){
cin>>a>>b>>c>>d;
x[m] = a;
t[m++] = seg(a,c,b,);
x[m] = c;
t[m++] = seg(a,c,d,-);
}
sort(x,x+m);
sort(t,t+m);
int nn = ;
for(int i = ;i < m;i++){
if(x[i]!=x[i-]) x[nn++] = x[i];
}
//for(int i = 0;i < nn;i ++)
// cout<<x[i]<<" ";
//cout<<endl;
double ret = ;
for(int i = ;i < m-;i ++){
int l = bin(t[i].l,nn,x);
int r = bin(t[i].r,nn,x)-;
//cout<<t[i].l<<" "<<t[i].r<<endl;
//cout<<"l: "<<l<<" r: "<<r<<endl;
if(l <= r) update(l,r,t[i].s,,nn-,);
//cout<<sum[1]<<endl;
ret += sum[] * (t[i+].h - t[i].h);
}
printf("Test case #%d\nTotal explored area: %.2lf\n\n",cas++ , ret);
memset(cnt,,sizeof(cnt));
memset(sum,,sizeof(sum));
}
return ;
}
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