# ZOJ 1610 Count the Colors（线段树lazy+暴力统计）

Count the Colors

Time Limit: 2 Seconds      Memory Limit: 65536 KB

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

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<string>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<bitset>
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
typedef long long LL;
typedef pair<int,int>pil;
const int maxn=8000+10;
int sum[maxn<<2];
int cnt,n;
int col[maxn],ans[maxn];
void pushdown(int rs)
{
if(sum[rs]!=⑴)
{
sum[rs<<1]=sum[rs<<1|1]=sum[rs];
sum[rs]=⑴;
}
}
void update(int x,int y,int c,int l,int r,int rs)
{
if(l>=x&&r<=y)
{
sum[rs]=c;
return ;
}
pushdown(rs);
int mid=(l+r)>>1;
if(x<=mid) update(x,y,c,l,mid,rs<<1);
if(y>mid) update(x,y,c,mid+1,r,rs<<1|1);
}
void solve(int l,int r,int rs)
{
if(l==r)
{
col[cnt++]=sum[rs];
return ;
}
pushdown(rs);
int mid=(l+r)>>1;
solve(l,mid,rs<<1);
solve(mid+1,r,rs<<1|1);
}
int main()
{
int l,r,x;
while(~scanf("%d",&n))
{
cnt=0;
CLEAR(sum,⑴);
CLEAR(ans,0);
int m=8000;
REP(i,n)
{
scanf("%d%d%d",&l,&r,&x);
update(l,r⑴,x,0,m,1);
}
solve(0,m,1);
REP(i,cnt)
{
int j=i+1;
if(col[i]!=⑴) ans[col[i]]++;
while(col[j]==col[i]&&j<cnt)
j++;
i=j⑴;
}
for(int i=0;i<=maxn;i++)//for色彩种类
if(ans[i]) printf("%d %d
",i,ans[i]);
puts("");
}
return 0;
}

1. 本站所有资源来源于用户上传和网络，如有侵权请邮件联系站长！
2. 分享目的仅供大家学习和交流，您必须在下载后24小时内删除！
3. 不得使用于非法商业用途，不得违反国家法律。否则后果自负！
4. 本站提供的源码、模板、插件等等其他资源，都不包含技术服务请大家谅解！
5. 如有链接无法下载、失效或广告，请联系管理员处理！
6. 本站资源售价只是赞助，收取费用仅维持本站的日常运营所需！
7. 本站源码并不保证全部能正常使用，仅供有技术基础的人学习研究，请谨慎下载
8. 如遇到加密压缩包，请使用WINRAR解压,如遇到无法解压的请联系管理员！