zoj 1610 Count the Colors 【暴力】

Count the Colors

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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

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Author: Standlove

Source: ZOJ Monthly, May 2003

思路：

        首先输入的是涂颜色的次数n，然后输入n行，每行输入三个数，分别代表其实坐标，终止坐标，还有是什么颜色；然后让你统计最终每一种颜色有多少块！

        首先，你将它的每一段的颜色都分成点，每个整数点所代表的颜色就有了，然后以颜色为数组col的下标，对应的值为这种颜色的快数，然后将col的非零值先输出下标，再输出对应的值，就OK了！

代码：

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;int n;int mp[8005];int col[8005];int main(){while(scanf("%d",&n)!=EOF){memset(mp,-1,sizeof(mp));//赋值为-1，不能赋值为0，因为颜色有0 memset(col,0,sizeof(col));for(int i=1;i<=n;i++){int a,b,c;scanf("%d%d%d",&a,&b,&c);for(int i=a;i<b;i++)//将每一个坐标点都图上相应的颜色 {mp[i]=c;}}int t=0,j=0;if(mp[0]!=-1)col[mp[0]]++;for(int i=1;i<8001;i++){if(mp[i]==-1)continue;if(mp[i]!=mp[i-1]){col[mp[i]]++;}}for(int i=0;i<8001;i++){if(col[i]!=0){printf("%d %d\n",i,col[i]);}}printf("\n");}return 0;}