题干：

It's graduated season, every students should leave something on the wall, so....they draw a lot of geometry shape with different color. 

When teacher come to see what happened, without getting angry, he was surprised by the talented achievement made by students. He found the wall full of color have a post-modern style so he want to have an in-depth research on it. 

To simplify the problem, we divide the wall into n*m (1 ≤ n ≤ 200, 1 ≤ m ≤ 50000) pixels, and we have got the order of coming students who drawing on the wall. We found that all students draw four kinds of geometry shapes in total that is Diamond, Circle, Rectangle and Triangle. When a student draw a shape in pixel (i, j) with color c (1 ≤ c ≤ 9), no matter it is covered before, it will be covered by color c. 

There are q (1 ≤ q ≤ 50000) students who have make a drawing one by one. And after q operation we want to know the amount of pixels covered by each color. 

Input

There are multiple test cases. 

In the first line of each test case contains three integers n, m, q. The next q lines each line contains a string at first indicating the geometry shape: 

* Circle: given xc, yc, r, c, and you should cover the pixels(x, y) which satisfied inequality (x - xc) 2 + (y - yc) 2 ≤ r 2 with color c; 
* Diamond: given xc, yc, r, c, and you should cover the pixels(x, y) which satisfied inequality abs(x - xc) + abs(y - yc) ≤ r with color c; 
* Rectangle: given xc, yc, l, w, c, and you should cover the pixels(x, y) which satisfied xc ≤ x ≤ xc+l-1, yc ≤ y ≤ yc+w-1 with color c; 
* Triangle: given xc, yc, w, c, W is the bottom length and is odd, the pixel(xc, yc) is the middle of the bottom. We define this triangle is isosceles and the height of this triangle is (w+1)/2, you should cover the correspond pixels with color c; 

Note: all shape should not draw out of the n*m wall! You can get more details from the sample and hint. (0 ≤ xc, x ≤ n-1, 0 ≤ yc, y ≤ m-1) 

Output

For each test case you should output nine integers indicating the amount of pixels covered by each color. 

Sample Input

8 8 4
Diamond 3 3 1 1
Triangle 4 4 3 2
Rectangle 1 1 2 2 3
Circle 6 6 2 4

Sample Output

4 4 4 11 0 0 0 0 0

Hint

The final distribution of different colors:
00000000
03300000
03310000
00111000
00022240
00002444
00004444
00000444

解题报告：

   其实这题建20棵线段树也可以轻松解决，，不过因为数据加强了好像zoj上会被T（因为内存给的很宽松）hdu上会MLE

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 2e5 + 5;
struct Node {
	char op[15];
	int x,y,l,w,c;
} node[MAX];
int f[MAX],ans[MAX];
int n,m,q;
bool vis[MAX];
int getf(int v) {
	return v == f[v] ? v : f[v] = getf(f[v]);
}
void merge(int u,int v) {
	int t1 = getf(u);
	int t2 = getf(v);
	if(t1!=t2) f[t2]=t1;//都归到左侧的t1上。。 
}
int main()
{	
	while(~scanf("%d%d%d",&n,&m,&q)) {
		for(int i = 1; i<=q; i++) {
			scanf("%s",node[i].op);
			if(node[i].op[0] == 'R') scanf("%d%d%d%d%d",&node[i].x,&node[i].y,&node[i].l,&node[i].w,&node[i].c);
			else scanf("%d%d%d%d",&node[i].x,&node[i].y,&node[i].l,&node[i].c);
		}
		memset(ans,0,sizeof ans);
		for(int r = 0; r<n; r++) {
			for(int c=0; c<m; c++) vis[c]=0,f[c]=c;
			for(int i = q; i>=1; i--) {
				int L,R,col = node[i].c;
				if(node[i].op[0] == 'C') {
					if(node[i].l < abs(r-node[i].x)) continue;
					int tmp = (int)sqrt(node[i].l*node[i].l - (r-node[i].x) * (r-node[i].x));//半径^2 - 竖直长度= 投影长度 
					L=node[i].y-tmp;
					R=node[i].y+tmp; 
				}
				if(node[i].op[0] == 'D') {
					if(node[i].l < abs(r-node[i].x)) continue;
					int tmp = node[i].l - abs(r-node[i].x);
					L=node[i].y-tmp;
					R=node[i].y+tmp;
				}
				if(node[i].op[0] == 'R') {
					if(r > node[i].x+node[i].l-1 || r < node[i].x) continue;
					L=node[i].y;
					R=node[i].y+node[i].w-1;
				}
				if(node[i].op[0] == 'T'){
					if(r-node[i].x >= (node[i].l+1)/2 || r<node[i].x) continue;
					int tmp = (node[i].l-1)/2 - (r-node[i].x);
					L=node[i].y-tmp;
					R=node[i].y+tmp;
				}
				L=max(0,L);R=min(R,m-1);
				int t1 = getf(L),t2;
				for(int c = R; c>=L; c=t2-1) {
					t2=getf(c);
					if(vis[t2]==0) ans[col]++,vis[t2]=1;//这里必须是标记t2，不能直接标记c。。因为你就不好查询是否被标记过了，，总之这里并查集都是用根节点进行操作，所以都转化成根节点就行了。 
					merge(t1,t2);
				}
			}
		}
		for(int i = 1; i<=9; i++) {
			printf("%d%c",ans[i],i==9?'\n':' ');
		}
	}
	return 0 ;
 }