ural 1147. Shaping Regions 几何

1147. Shaping Regions

Time limit: 0.5 second
Memory limit: 64 MB

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.

Input

The order of the input lines dictates the order of laying down the rectangles. The first input line is a rectangle “on the bottom”. First line contains A, B and N, space separated (1 ≤ A, B ≤ 10000). Lines 2, …, N + 1 contain five integers each: 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.

Output

The output 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.

Sample

inputoutput

20 20 32 2 18 18 20 8 19 19 38 0 10 19 4

1 912 843 1874 38

题意： 对照案例， 有20*20个格子，  有三次涂颜色的操作， 接下来涂颜色，输入坐标和颜色  llx, lly, urx, ury, color。  后涂会覆盖之前涂的，最后输出的是个个颜色的面积，没涂颜色的算颜色1。

做法： 几何，从第一个开始， 处理第一个的时候不断找其上层的 有没有覆盖他，  有的话去掉覆盖的部分继续 处理剩下的部分。

#include<stdio.h>#include<string.h>#define max(a,b) a>b?a:b#define min(a,b) a<b?a:bstruct rectangle {int x1,y1,x2,y2; int c; };int cal(int x1,int y1,int x2,int y2){return (x2-x1)*(y2-y1);} rectangle rect[1100];int area_num[2600];//1-2500 int area_cal(int x1,int y1,int x2,int y2,const int& k,const int& n){int kk;for(int i=k+1;i<=n+1;i++){if(i==n+1)return cal(x1,y1,x2,y2);if(x1>=rect[i].x2||x2<=rect[i].x1||y1>=rect[i].y2||y2<=rect[i].y1)//没有覆盖continue; else{kk=i;break;}}int sum=0;if(x1<rect[kk].x1){sum+=area_cal(x1,y1,rect[kk].x1,y2,kk,n);x1=rect[kk].x1;}if(x2>rect[kk].x2){ sum+=area_cal(rect[kk].x2,y1,x2,y2,kk,n); x2=rect[kk].x2;}if(y1<rect[kk].y1){ sum+=area_cal(x1,y1,x2,rect[kk].y1,kk,n); y1=rect[kk].y1;}if(y2>rect[kk].y2){ sum+=area_cal(x1,rect[kk].y2,x2,y2,kk,n); y2=rect[kk].y2;} return sum;}int main(){int a,b,n;while(scanf("%d%d%d",&a,&b,&n)!=EOF){memset(area_num,0,sizeof area_num);area_num[1]=a*b;for(int i=1;i<=n;i++){scanf("%d%d%d%d%d",&rect[i].x1,&rect[i].y1,&rect[i].x2,&rect[i].y2,&rect[i].c);}for(int i=n;i>=1;i--){int tem=area_cal(rect[i].x1,rect[i].y1,rect[i].x2,rect[i].y2,i,n);area_num[rect[i].c]+=tem;area_num[1]-=tem;}for(int i=1;i<=2500;i++){if(area_num[i])printf("%d %d\n",i,area_num[i]); } } return 0; }/*20 20 32 2 18 18 20 8 19 19 38 0 10 19 44 4 21 1 3 3 32 2 4 4 2 右上4 4 21 1 3 3 30 2 2 4 2 左上 1 912 843 1874 38*/