Draw a Mess
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 864 Accepted Submission(s): 180
Problem Description
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 ≤ r2 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
一开始用线段树做，hdu上弹MLE，然后TLE，但是zojAC。最后看了别人怎么做的，竟然用并查集，又是新技能呀！get！
思路：
乍一看，行只有200，比列少很多，所以逐行处理每个图形覆盖到该行的区间。之后的图形覆盖前面的图形，所以，我们在这里从后往前处理。
现在，我们来看看每一行是怎么处理的：
如下图，顶层模型是我从线段的右端开始涂色，当涂到已经涂过的格子时，我直接跳到下一个未涂色的格子，这样就减少了遍历涂过色格子的时间。而如何跳到我们需要的格子呢？就用到了并查集的东西了。
#include <string>
#include <cmath>
#include <iostream>
#include <cstdio>
#include <vector>
using namespace std;
const double eps = 1e-8;
const int maxn = 210;
const int maxm = 50010;
int m , n , q , color[15] , father[maxm] , vis[maxm];
struct Node{
char op;
int xc , yc , r , l , w,c;
Node(char OP = 'x' , int XC = 0, int YC = 0, int R = 0, int L = 0 , int W = 0, int C = 0){
op = OP , xc = XC , yc = YC , l = L , r = R , w = W , c = C;
}
};
vector<Node> v;
int findFather(int x){
if(father[x] != x) return father[x] = findFather(father[x]);
return father[x];
}
inline bool getCircle(int pos,int xc,int yc,int r,int &L,int &R) {
int tmp = r*r-(pos-xc)*(pos-xc);
if(tmp<0) return false;
L = int(-sqrt(tmp*1.0)+yc+1.0-eps);
//L = max(0,L);
if(0 > L) L = 0;
R = int(sqrt(tmp*1.0)+yc);
//R = min(m-1,R);
if(R > m-1) R = m-1;
if(L>R) return false;
return true;
}
inline bool getRect(int pos,int xc,int yc,int l,int w,int &L,int &R) {
if(pos<xc || pos >= xc+l) return false;
else {
L = yc;
//L = max(0,L);
if(0 > L) L = 0;
R = yc+w-1;
//R = min(m-1,R);
if(R > m-1) R = m-1;
if(L>R) return false;
return true;
}
}
inline bool getDiamond(int pos,int xc,int yc,int r , int &L , int &R) {
int tmp = r - abs(xc-pos);
if(tmp < 0) return false;
L = yc-tmp;
//L = max(0,L);
if(0 > L) L = 0;
R = yc+tmp;
//R = min(m-1,R);
if(R > m-1) R = m-1;
if(L>R) return false;
return true;
}
inline bool getTriangle(int pos , int xc , int yc , int w , int &L , int &R){
if(pos < xc) return false;
L = yc-w/2+pos-xc;
R = yc+w/2-(pos-xc);
//L = max(L , 0);
if(0 > L) L = 0;
//R = min(R , m-1);
if(R > m-1) R = m-1;
if(L>R) return false;
return true;
}
void initial(){
for(int i = 0; i < 15; i++) color[i] = 0;
v.clear();
for(int i = 0; i < maxm; i++) father[i] = i , vis[i] = 0;
}
void computing(){
char shape[20];
int xc , yc , w , r , c , l , L , R;
while(q--){
scanf("%s" , shape);
if(shape[0] == 'C'){
scanf("%d%d%d%d" , &xc , &yc , &r , &c);
}
if(shape[0] == 'D'){
scanf("%d%d%d%d" , &xc , &yc , &r , &c);
}
if(shape[0] == 'R'){
scanf("%d%d%d%d%d" ,&xc ,&yc ,&l ,&w ,&c);
}
if(shape[0] == 'T'){
scanf("%d%d%d%d" ,&xc,&yc,&w,&c);
}
v.push_back(Node(shape[0] , xc , yc , r , l , w , c));
}
int fl , fr;
for(int i = 0; i < n; i++){
//cout << i << ": ";
for(int j = 0; j < m; j++) father[j] = j , vis[j] = 0;
for(int k = v.size()-1; k >= 0; k--){
if(v[k].op == 'C' && !getCircle(i , v[k].xc , v[k].yc , v[k].r , L , R))continue;
if(v[k].op == 'D' && !getDiamond(i , v[k].xc , v[k].yc , v[k].r , L , R)) continue;
if(v[k].op == 'R' && !getRect(i , v[k].xc , v[k].yc , v[k].l , v[k].w , L , R)) continue;
if(v[k].op == 'T' && !getTriangle(i , v[k].xc , v[k].yc , v[k].w , L , R)) continue;
c = v[k].c;
//cout << L << " " << R << " " << c << endl;
int fl = findFather(L);
while(R>=L){
if(vis[R] == 0){
vis[R] = 1;
color[c]++;
}
fr = findFather(R);
R = min(fr-1 , R-1);
if(fr > fl)father[fr] = fl;
}
}
}
printf("%d" , color[1]);
for(int i = 2; i <= 9; i++) printf(" %d" , color[i]);
printf("\n");
}
int main(){
while(~scanf("%d%d%d" , &n , &m , &q)){
initial();
computing();
}
return 0;
}