zoj 1610 Count the Colors 线段树区间更新——染色问题

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
题目大意：给了每一线段的颜色，存在颜色覆盖，求表面上看能看到的颜色种类和各种颜色的段数；
链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1610
代码：

#include <iostream>
#include <stdio.h>
#include <string.h>
#define N 8005
using namespace std;
int temp,col
;
struct Tree
{
int l,r;
int color;
}tree[N*4];
void build(int l,int r,int root)
{
tree[root].l=l;
tree[root].r=r;
tree[root].color=-1;
if(l+1==r) return;
int mid=(l+r)>>1;
build(l,mid,root<<1);
build(mid,r,(root<<1|1));
}
void insert(int root,int l,int r,int c)
{
if(l==r) return;
if(tree[root].color==c) return;
if(tree[root].l>=l&&tree[root].r<=r)
{
tree[root].color=c;
return;
}
if(tree[root].color>=0)
{
tree[root<<1].color=tree[root<<1|1].color=tree[root].color;
tree[root].color=-2;
}
int mid=(tree[root].l+tree[root].r)/2;
if(mid<=l) insert(root<<1|1,l,r,c);
else if(mid>=r)  insert(root<<1,l,r,c);
else
{
insert(root<<1,l,mid,c);
insert(root<<1|1,mid,r,c);
}
tree[root].color=-2;
}
void search(int root)
{
if(tree[root].color==-1)
{
temp=-1;
return;
}
if(tree[root].color!=-2)
{
if(temp!=tree[root].color)
{
col[tree[root].color]++;
temp=tree[root].color;
}
return;
}
if(tree[root].l+1!=tree[root].r)
{
search(root<<1);
search(root<<1|1);
}
}
int main()
{
int n,a,b,c;
while(scanf("%d",&n)!=EOF)
{
build(0,8000,1);
int maxn=0;
for(int i=1;i<=n;i++)
{
scanf("%d%d%d",&a,&b,&c);
insert(1,a,b,c);
if(maxn<c) maxn=c;
}
memset(col,0,sizeof(col));
search(1);
for(int i=0;i<=maxn;i++)
{
if(col[i]) printf("%d %d\n",i,col[i]);
}
printf("\n");
}
return 0;
}