zoj1610Count the Colors【暴力 or 线段树】

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本文详细解析了ZOJ Monthly May 2003中的计数不同颜色段问题，通过输入的彩色段落集合，采用高效的算法策略，实现了颜色段的计数，并输出可见颜色及其数量。通过实例输入，展示了算法的应用过程及输出结果。

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ZOJ - 1610

Count the Colors

Time Limit: 2000MS

Memory Limit: 65536KB

64bit IO Format: %lld & %llu

Submit Status

Description

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

Source

ZOJ Monthly, May 2003

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int sum[8010];
int vis[8010];
int main()
{
	int n,i,j,k;
	while(scanf("%d",&n)!=EOF){
		memset(sum,-1,sizeof(sum));
		int a,b,c;
		for(i=0;i<n;++i){
			scanf("%d%d%d",&a,&b,&c);
			for(j=a+1;j<=b;++j){
				sum[j]=c;
			}
		}
		memset(vis,0,sizeof(vis));
		for(i=1;i<=8001;++i){
			if(sum[i]!=sum[i-1]){
				if(sum[i-1]!=-1){
					vis[sum[i-1]]++;
				}
			}
		}
		for(i=0;i<=8000;++i){
			if(vis[i]){
				printf("%d %d\n",i,vis[i]);
			}
		}
		printf("\n");
	}
	return 0;
}