解題報告
題意:
給定的矩形裡面有镂空的矩陣,求矩陣面積並。
思路:
直接把一個圖形拆成4個矩形,進行面積並。
掃描線+線段樹
#include #include#include #include #define LL __int64 using namespace std; struct Seg { int lx,rx,h,v; friend bool operator < (Seg a,Seg b) { return a.h r||qr >1; update(rt<<1,l,mid,ql,qr,v); update(rt<<1|1,mid+1,r,ql,qr,v); push_up(rt,l,r); } int main() { int n,i,j,x1,y1,x2,y2,x3,y3,x4,y4; while(~scanf("%d",&n)) { if(!n)break; int m=0; memset(lz,0,sizeof(lz)); memset(sum,0,sizeof(sum)); for(i=0; i
Posters
Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4585 Accepted Submission(s): 1029
Problem Description Ted has a new house with a huge window. In this big summer, Ted decides to decorate the window with some posters to prevent the glare outside. All things that Ted can find are rectangle posters.
However, Ted is such a picky guy that in every poster he finds something ugly. So before he pastes a poster on the window, he cuts a rectangular hole on that poster to remove the ugly part. Ted is also a careless guy so that some of the pasted posters may overlap when he pastes them on the window.
Ted wants to know the total area of the window covered by posters. Now it is your job to figure it out.
To make your job easier, we assume that the window is a rectangle located in a rectangular coordinate system. The window’s bottom-left corner is at position (0, 0) and top-right corner is at position (50000, 50000). The edges of the window, the edges of the posters and the edges of the holes on the posters are all parallel with the coordinate axes.
Input The input contains several test cases. For each test case, the first line contains a single integer N (0
The input ends with a line of single zero.
Output For each test case, output a single line with the total area of window covered by posters.
Sample Input2 0 0 10 10 1 1 9 9 2 2 8 8 3 3 7 7 0
Sample Output56
Source 2009 Asia Ningbo Regional Contest Hosted by NIT