F - Count the Colors（线段树）

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.

输入：

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.

输出：

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：这个题目和D - Mayor’s posters这个题目很相似，只是这个不需要映射，而那个题目需要

要注意的就是：我们的线段树只能存点，我们不能把一个[1,4]区间用四个点去存放，因为实际上区间长度为3，如果我们用[1,4]内所有点去存就表示长度为三

解决方法：

我们只需要再给出的区间在其左边界加一，或是在右边界减一从而使其区间中的点变得和区间长度一样

上代码；

#include<cstdio>

#include<cmath>

#include<algorithm>

#include<cstring>

using namespace std;

const int maxn=8005;

struct node

{

	int l,r,sum;

}a[maxn<<2];

int x1[maxn],x2[maxn],c[maxn];

int mark[maxn];

int sum[maxn];

int lenn; 

void build(int o,int l,int r)

{

	a[o].l=l,a[o].r=r,a[o].sum=-1;

	int ls=o<<1,rs=o<<1|1,mid=(l+r>>1);

	if(l==r)

	{

		a[o].sum=-1;

		return ;

	}

	build(ls,l,mid);

	build(rs,mid+1,r);

}

void pushdown(int o)

{

	   if(a[o].sum!=-1)

	   {

	   		a[o<<1].sum=a[o].sum;

	   		a[o<<1|1].sum=a[o].sum;

	   		a[o].sum=-1;

	   }

}

void update(int o,int l,int r,int c)

{

	int ls=o<<1,rs=o<<1|1,mid=(a[o].l+a[o].r)>>1;

	if(l<=a[o].l&&a[o].r<=r)

	{

		a[o].sum=c;

		return ;

	}

	pushdown(o);

	if(l<=mid) update(ls,l,r,c);

	if(r>mid) update(rs,l,r,c);

}

void query(int o,int l,int r)

{

	int ls=o<<1,rs=o<<1|1,mid=(a[o].l+a[o].r)>>1;

	if(a[o].l==a[o].r)

	{

		mark[lenn++]=a[o].sum;

		return ;

	}

	pushdown(o);

	if(l<=mid) query(ls,l,r);

	if(r>mid) query(rs,l,r);

}

int main()

{

	int n;

	while(~scanf("%d",&n))

	{

		int maxx=0;

		for(int i=0;i<n;i++)

		{

			scanf("%d%d%d",&x1[i],&x2[i],&c[i]);

			int x=max(x1[i]+1,x2[i]);

			maxx=max(maxx,x);

		}

		memset(mark,-1,sizeof(mark));

		build(1,1,maxx);

		for(int i=0;i<n;i++)	update(1,x1[i]+1,x2[i],c[i]);

		lenn=0;

		query(1,1,maxx);

		int x;

		memset(sum,0,sizeof(sum));

		for(int i=0;i<lenn;)

		{

			if(mark[i]==-1)	

			{

				i++;

				continue;

			}

			x=mark[i];

			while(x==mark[++i]&&i<lenn) ;

			sum[x]++;

		}

		for(int i=0;i<=maxx+10;i++)

		{

			if(sum[i]!=0)

				printf("%d %d\n",i,sum[i]);

		}

		printf("\n");  /注意不要忘了最后还有一个换行

	}

}