Count the Colors（线段树 + 延迟标记）

Count the Colors

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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Painting some colored segments on a line, some previously painted segments may be covered by some the subsequent ones.

Your task is counting the segments of different colors you can see at last.

Input

The first line of each data set contains exactly one integer n, 1 <= n <= 8000, equal to the number of colored segments.

Each of the following n lines consists of exactly 3 nonnegative integers separated by single spaces:

x1 x2 c

x1 and x2 indicate the left endpoint and right endpoint of the segment, c indicates the color of the segment.

All the numbers are in the range [0, 8000], and they are all integers.

Input may contain several data set, process to the end of file.

Output

Each line of the output should contain a color index that can be seen from the top, following the count of the segments of this color, they should be printed according to the color index.

If some color can't be seen, you shouldn't print it.

Print a blank line after every dataset.

Sample Input

5
0 4 4
0 3 1
3 4 2
0 2 2
0 2 3
4
0 1 1
3 4 1
1 3 2
1 3 1
6
0 1 0
1 2 1
2 3 1
1 2 0
2 3 0
1 2 1

Sample Output

1 1
2 1
3 1

1 1

0 2
1 1

 

   题意：

   有N（1到8000）种颜色，给出N行a,b,c，表示a到b这个区间填充c颜色，N次输入的颜色可以覆盖前面的颜色，最后输出能看见的颜色中，一共有几段（颜色要按从小到大输出）。

 

   思路：

   用线段树延迟标记。

 

   AC：

#include<cstdio>
#include<string.h>
#include<map>
#define MAX (8000+5)
#define INT 10000
using namespace std;
typedef struct
{
	int l;   //左边界
	int r;   //右边界
	int co;  //颜色域
	int ad;  //标记域
}node;
node no[MAX*3];
int col[MAX],from[MAX],to[MAX];
int fin[MAX];  //fin用来保存这条线段最后的颜色


void build(int from,int to,int i)
{
	int mid=(from+to)/2;
	no[i].l=from;
	no[i].r=to;
	no[i].ad=-1;
	no[i].co=-1;
	if(to-from==1) return;
	build(from,mid,2*i);
	build(mid,to,2*i+1);
}

void add(int from,int to,int i,int c)
{
	int mid=(no[i].l+no[i].r)/2;
	if(from==no[i].l&&to==no[i].r)
	{
		no[i].ad=c;
		return;
	}
	else
	{
		if(no[i].ad!=-1)
		{
			no[i].co=no[i].ad;
			no[2*i].ad=no[i].ad;
			no[2*i+1].ad=no[i].ad;
			no[i].ad=-1;
		}
		if(from>=mid) add(from,to,2*i+1,c);
		if(to<=mid)   add(from,to,2*i,c);
		if(from<mid&&to>mid)
		{
			add(from,mid,2*i,c);
			add(mid,to,2*i+1,c);
		}
	}
}

void find(int i)
{
	int mid=(no[i].l+no[i].r)/2;
	if(no[i].r-no[i].l==1)
	{
		if(no[i].ad!=-1) no[i].co=no[i].ad;
		fin[no[i].l]=no[i].co;
	}
	else
	{
		if(no[i].ad!=-1)
		{
			no[i].co=no[i].ad;
			no[2*i].ad=no[i].ad;
			no[2*i+1].ad=no[i].ad;
			no[i].ad=-1;
		}
		find(2*i);
		find(2*i+1);
	}
}

int main()
{
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		int ma=-1,mi=INT;
		memset(fin,-1,sizeof(fin));
		for(int i=1;i<=n;i++)
		{
			scanf("%d%d%d",&from[i],&to[i],&col[i]);
			if(from[i]<mi) mi=from[i];
			if(to[i]>ma)   ma=to[i];
		}
//建树
		build(mi,ma,1);
//添加颜色
		for(int i=1;i<=n;i++)
			add(from[i],to[i],1,col[i]);
//查找颜色段
		find(1);
//用map关联容器存储最后结果
		map<int,int> m;
		m.clear();
		for(int i=mi;i<=ma-1;i++)
			{
				if(fin[i]==-1) continue;
				if(i==mi&&fin[mi]!=-1)      m[fin[i]]++;
				if(i>mi&&fin[i]!=fin[i-1])  m[fin[i]]++;
			}
//输出结果
		map<int,int>::iterator it;
		for(it=m.begin();it!=m.end();it++)
		 printf("%d %d\n",it->first,it->second);
		printf("\n");
	}
	return 0;
}

 

    总结：

    1.线段树的区间要想清楚如何分法；

    2.查询完后，想清楚要如何判断有几段；