HDU4419 Colourful Rectangle(矩形面积并+线段树)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4419

解析：跟一般矩形面积并有一点的区别，就是每条线的状态要用一个数组来维护，这里的depth[ i ] 表示当前状态第i中颜色的已覆盖的长度，三种基本颜色可以分别用1，2，4表示，然后通过二进制运算得到其他颜色的标号，其他也没什么了。

参考代码：

#include
#include
#include
#include
#include
#include
using namespace std;

#define clr(arr,val) memset(arr,val,sizeof(arr))
const int M = 20500;

enum In{lt,rt};
int Out[8] = {0,1,2,4,3,5,6,7};
int n,k;
int num_y[M*2];
struct rectangle{
	int colour;
	int x1,y1,x2,y2;
}rec[M];

struct SegTree{
	int count[8];
	int L,R;
	long long depth[8];
}Tree[M*8];

void Update(int);
void Build(int root,int L,int R){
	clr(Tree[root].count,0);
	Tree[root].L = L;
	Tree[root].R = R;
	if(L == R - 1)  return ;
	Build(root<<1,L,(L+R)/2);
	Build((root<<1)+1,(L+R)/2,R);
}
void Insert(int root,int L,int R,int colour){
	if(num_y[ Tree[root].L ] >= L && num_y[ Tree[root].R ] <= R){
		Tree[root].count[colour]++;
		Update(root);
	}
	else {
		if(L < num_y[ (Tree[root].L+Tree[root].R)/2 ]){
			Insert(root<<1,L,R,colour);
			Update(root);
		}
		if(R > num_y[ (Tree[root].L+Tree[root].R)/2 ]){
			Insert((root<<1)+1,L,R,colour);
			Update(root);
		}
	}
}
void Delete(int root,int L,int R,int colour){
	if(num_y[ Tree[root].L ] >= L && num_y[ Tree[root].R ] <= R){
		Tree[root].count[colour]--;
		Update(root);
	}
	else {
		if(L < num_y[ (Tree[root].L+Tree[root].R)/2 ]){
			Delete(root<<1,L,R,colour);
			Update(root);
		}
		if(R > num_y[ (Tree[root].L+Tree[root].R)/2 ]){
			Delete((root<<1)+1,L,R,colour);
			Update(root);
		}
	}
}
void Update(int root){
	clr(Tree[root].depth,0);
	int colour = (Tree[root].count[1]>0?1:0) | (Tree[root].count[2]>0?2:0) | (Tree[root].count[4]>0?4:0);
	if(colour > 0){
		Tree[root].depth[colour] += num_y[ Tree[root].R ] - num_y[ Tree[root].L ];
		for(int i = 1;i != 8;++i){
			if((i | colour) != colour){
				int tmp = Tree[root<<1].depth[i] + Tree[(root<<1)+1].depth[i];
				Tree[root].depth[colour] -= tmp;
				Tree[root].depth[i|colour] += tmp;
			}
		}
	}
	else if(Tree[root].L != Tree[root].R - 1) {
			for(int i = 1;i != 8;++i)
				Tree[root].depth[i] = Tree[root<<1].depth[i] + Tree[(root<<1)+1].depth[i];
	}
}
struct operation{
	In in;
	int x,lower_y,high_y,colour;
}Event[M*4];

void input(){
	cin>>n;
	char ch;
	for(int i = 0;i != n;++i){
		cin>>ch>>rec[i].x1>>rec[i].y1>>rec[i].x2>>rec[i].y2;
		if(ch == 'R') rec[i].colour = 1;
		if(ch == 'G') rec[i].colour = 2;
		if(ch == 'B') rec[i].colour = 4;
	}
}
void Hash(){
	for(int i = 0;i != n;++i){
		num_y[i*2] = rec[i].y1;
		num_y[i*2+1] = rec[i].y2;

		Event[i*2].in = lt;
		Event[i*2].x = rec[i].x1;
		Event[i*2].lower_y = rec[i].y1;
		Event[i*2].high_y = rec[i].y2;
		Event[i*2].colour = rec[i].colour;

		Event[i*2+1].in = rt;
		Event[i*2+1].x = rec[i].x2;
		Event[i*2+1].lower_y = rec[i].y1;
		Event[i*2+1].high_y = rec[i].y2;
		Event[i*2+1].colour = rec[i].colour;
	}
	sort(num_y,num_y+n*2);
}
bool cmp(operation a,operation b){ return a.x < b.x; }
void solve(){
	input();
	Hash();
	Build(1,0,n*2-1);
	sort(Event,Event+n*2,cmp);
	long long ans[8],depth[8];
	clr(ans,0);
	clr(depth,0);
	for(int i = 0;i < n*2;++i){
		if(Event[i].in == lt) 
			Insert(1,Event[i].lower_y,Event[i].high_y,Event[i].colour);
		else
			Delete(1,Event[i].lower_y,Event[i].high_y,Event[i].colour);
		for(int j = 1;j != 8;++j){
			ans[j] += (long long)depth[j] * (long long)(i?(Event[i].x-Event[i-1].x):0);
		}
		memcpy(depth,Tree[1].depth,sizeof(Tree[1].depth));
	}
	cout<<"Case "<<++k<<":"<>T;
	while(T--){
		solve();
	}
	return 0;
}