Count the Colors

ZOJ Problem Set - 1610

​​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

Author:

Standlove

Source: ZOJ Monthly, May 2003

评测连接：​​/news/upload/ueditor/image/202209/1wbbtjaatkv.cn style="color: rgb(89, 89, 89); margin: 0px; padding: 0px; background: none 0% 0% / auto repeat scroll padding-box border-box rgba(0, 0, 0, 0);">#include<cstdio>
#include<algorithm>
#include<cstring>
#define maxn (8000+100)
/*
#define maxn 8000+100---->maxn*4==8000+100*4
#define maxn (8000+100)---->maxn*4==(8000+100)*4
*/
using namespace std;

int n,last=-1,lasts=-1,cmax=0,dmax=0;
int x[maxn],y[maxn],c[maxn];
int sum[maxn],color[maxn*4];

void init()
{
freopen("color.in","r",stdin);
freopen("color.out","w",stdout);
}

void paint(int p,int pl,int pr,int l,int r,int s)
{
if(l>r)return;
if(l>pr || r<pl)return;
if(l<=pl && pr<=r){color[p]=s;return;}
int m=(pl+pr)>>1,k=p<<1;
if(color[p]!=-2)//单需要对区间p进行一定改动时，才把p的信息传给子节点，俗称懒标记
{
color[k]=color[k+1]=color[p];
color[p]=-2;
}
if(l<=m)paint(k,pl,m,l,r,s);
if(r>m)paint(k+1,m+1,pr,l,r,s);
}

void find(int p,int l,int r)
{
if(l>r)return;
if(color[p]>=-1)
{ //即使color[p]==-1,只是sum[-1]加1
if(color[p]!=last)++sum[color[p]],last=color[p];
return;
}
if(l==r)return;
if(color[p]!=-2)//这里也需要向下传标记，因为可能出现用到区间p1，但p1的父亲并未把节点
color[p<<1]=color[(p<<1)+1]=color[p];//信息向下传达，也就是p1线段应为颜色i，但查找时color[p1]却是等于-1的情况
int m=(l+r)>>1,k=p<<1;
find(k,l,m),find(k+1,m+1,r);
}

void work()
{
memset(sum,0,sizeof(sum));
memset(color,-1,sizeof(color));
int i;
for(i=1;i<=n;i++)
{//用点i+1代表线段[i,i+1];
scanf("%d%d%d",&x[i],&y[i],&c[i]);x[i]++;
cmax=max(cmax,c[i]); dmax=max(dmax,y[i]);
}
//染色
for(i=1;i<=n;i++)paint(1,1,dmax,x[i],y[i],c[i]);
find(1,1,dmax);//记录每种颜色有多少段
for(i=0;i<=cmax;i++)if(sum[i])printf("%d %d\n",i,sum[i]);
printf("\n");
}

int main()
{
init();
while(scanf("%d",&n)!=EOF)work();
return 0;
}