Shaping Regions
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.
PROGRAM NAME: rect1
INPUT FORMAT
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. |
SAMPLE INPUT (file rect1.in)
20 20 3 2 2 18 18 2 0 8 19 19 3 8 0 10 19 4
INPUT EXPLANATION
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 at 8,0 and a 4 at 8,1 but NOT a 4 at 8,2 since this diagram can't capture what would be shown on graph paper).
OUTPUT FORMAT
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.
SAMPLE OUTPUT (file rect1.out)
1 91 2 84 3 187 4 38
————————————————————————————————————————————题解
和前面那个窗口题有什么区别……？？？
1 /* 2 ID: ivorysi 3 LANG: C++ 4 PROG: rect1 5 */ 6 #include <iostream> 7 #include <cstdio> 8 #include <cstring> 9 #include <queue> 10 #include <set> 11 #include <vector> 12 #include <algorithm> 13 #define siji(i,x,y) for(int i=(x);i<=(y);++i) 14 #define gongzi(j,x,y) for(int j=(x);j>=(y);--j) 15 #define xiaosiji(i,x,y) for(int i=(x);i<(y);++i) 16 #define sigongzi(j,x,y) for(int j=(x);j>(y);--j) 17 #define inf 0x5f5f5f5f 18 #define ivorysi 19 #define mo 97797977 20 #define hash 974711 21 #define base 47 22 #define fi first 23 #define se second 24 #define pii pair<int,int> 25 #define esp 1e-8 26 typedef long long ll; 27 using namespace std; 28 int n,col[10005],an[2505]; 29 struct data { 30 int lx,ly,rx,ry; 31 }squ[10005]; 32 int areas(data &t) { 33 return (t.rx-t.lx+1)*(t.ry-t.ly+1); 34 } 35 void init() { 36 scanf("%d%d%d",&squ[0].rx,&squ[0].ry,&n); 37 --squ[0].rx;--squ[0].ry; 38 siji(i,1,n) { 39 scanf("%d%d%d%d",&squ[i].lx,&squ[i].ly,&squ[i].rx,&squ[i].ry); 40 --squ[i].rx; 41 --squ[i].ry; 42 scanf("%d",&col[i]); 43 } 44 col[0]=1; 45 } 46 int dfs(int k,data q) { 47 if(q.rx<q.lx || q.ry<q.ly) return 0; 48 49 50 int ans=0; 51 while((q.lx>squ[k].rx || q.rx<squ[k].lx || q.ly>squ[k].ry || q.ry<squ[k].ly )&&k<=n) ++k; 52 if(k>n) return areas(q); 53 if(q.lx<squ[k].lx) 54 ans+=dfs(k+1,(data){q.lx,max(q.ly,squ[k].ly),squ[k].lx-1,min(squ[k].ry,q.ry)}); 55 if(q.rx>squ[k].rx) 56 ans+=dfs(k+1,(data){squ[k].rx+1,max(q.ly,squ[k].ly),q.rx,min(squ[k].ry,q.ry)}); 57 if(q.ly<squ[k].ly) 58 ans+=dfs(k+1,(data){q.lx,q.ly,q.rx,squ[k].ly-1}); 59 if(q.ry>squ[k].ry) 60 ans+=dfs(k+1,(data){q.lx,squ[k].ry+1,q.rx,q.ry}); 61 return ans; 62 } 63 void solve() { 64 init(); 65 siji(i,0,n) 66 an[col[i]]+=dfs(i+1,squ[i]); 67 siji(i,1,2500) if(an[i]!=0) printf("%d %d ",i,an[i]); 68 } 69 int main(int argc, char const *argv[]) 70 { 71 #ifdef ivorysi 72 freopen("rect1.in","r",stdin); 73 freopen("rect1.out","w",stdout); 74 #else 75 freopen("f1.in","r",stdin); 76 #endif 77 solve(); 78 return 0; 79 }