N opaque rectangles (1 <= N <= 1000) of various colors are placed on a white sheet of paper whose size is A wide by B long. The rectangles are put with their sides parallel to the sheet's borders. All rectangles fall within the borders of the sheet so that different figures of different colors will be seen.
The coordinate system has its origin (0,0) at the sheet's lower left corner with axes parallel to the sheet's borders.
The order of the input lines dictates the order of laying down the rectangles. The first input line is a rectangle "on the bottom".
Line 1: | A, B, and N, space separated (1 <= A,B <= 10,000) |
Lines 2-N+1: | Five integers: llx, lly, urx, ury, color: the lower left coordinates and upper right coordinates of the rectangle whose color is `color' (1 <= color <= 2500) to be placed on the white sheet. The color 1 is the same color of white as the sheet upon which the rectangles are placed. |
20 20 3 2 2 18 18 2 0 8 19 19 3 8 0 10 19 4
Note that the rectangle delineated by 0,0 and 2,2 is two units wide and two high. Here's a schematic diagram of the input:
11111111111111111111 33333333443333333331 33333333443333333331 33333333443333333331 33333333443333333331 33333333443333333331 33333333443333333331 33333333443333333331 33333333443333333331 33333333443333333331 33333333443333333331 33333333443333333331 11222222442222222211 11222222442222222211 11222222442222222211 11222222442222222211 11222222442222222211 11222222442222222211 11111111441111111111 11111111441111111111
The '4's at 8,0 to 10,19 are only two wide, not three (i.e., the grid contains a 4 and 8,0 and a 4 and 8,1 but NOT a 4 and 8,2 since this diagram can't capture what would be shown on graph paper).
The output file should contain a list of all the colors that can be seen along with the total area of each color that can be seen (even if the regions of color are disjoint), ordered by increasing color. Do not display colors with no area.
1 91 2 84 3 187 4 38
题意:给定一个大小为 a x b 的矩阵,在第一象限内。起初每个整点的颜色都是 1 ,现在用 n 个不同颜色的矩形去覆盖,求 n 次覆盖后每种颜色有多少。
分析:漂浮法,想象一下一共有 n 层,从底向上依次上浮,有重叠的去掉。
/* ID: dizzy_l1 LANG: C++ TASK: rect1 */ #include<iostream> #include<cstring> #include<cstdio> #define MAXC 2501 #define MAXN 1001 using namespace std; struct rect1 { int x1,y1,x2,y2,c; }r[MAXN]; int ans[MAXC],n; void Search(int x1,int y1,int x2,int y2,int c,int k) { if(k==n) { ans[c]+=(x2-x1)*(y2-y1); return ; } if(x2<=r[k].x1||x1>=r[k].x2||y2<=r[k].y1||y1>=r[k].y2) { Search(x1,y1,x2,y2,c,k+1); return ; } if(x1<=r[k].x1) { Search(x1,y1,r[k].x1,y2,c,k+1); x1=r[k].x1; } if(x2>=r[k].x2) { Search(r[k].x2,y1,x2,y2,c,k+1); x2=r[k].x2; } if(y1<=r[k].y1) { Search(x1,y1,x2,r[k].y1,c,k+1); y1=r[k].y1; } if(y2>=r[k].y2) { Search(x1,r[k].y2,x2,y2,c,k+1); } } int main() { freopen("rect1.in","r",stdin); freopen("rect1.out","w",stdout); int a,b,i; while(scanf("%d %d %d",&a,&b,&n)==3) { memset(ans,0,sizeof(ans)); for(i=0;i<n;i++) scanf("%d%d%d%d%d",&r[i].x1,&r[i].y1,&r[i].x2,&r[i].y2,&r[i].c); for(i=0;i<n;i++) { Search(r[i].x1,r[i].y1,r[i].x2,r[i].y2,r[i].c,i+1); } ans[1]=a*b; for(i=2;i<MAXC;i++) ans[1]-=ans[i]; for(i=1;i<MAXC;i++) { if(ans[i]) printf("%d %d\n",i,ans[i]); } } return 0; }