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Description
Your task is counting the segments of different colors you can see at last.
Input
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
If some color can‘t be seen, you shouldn‘t print it.
Print a blank line after every dataset.
Sample Input
Sample Output
1 1
0 2
1 1
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
#include<cstdlib>
using namespace std;
#define N 8001
int n,color[N],cnt[N];
int main()
{
while(scanf("%d",&n)!=EOF)
{
int x,y,z,i,j,maxx=0;
memset(color,0,sizeof(color));
memset(cnt,0,sizeof(cnt));
for(i=0;i<n;i++)
{
scanf("%d%d%d",&x,&y,&z);
for(j=x;j<y;j++)
color[j]=z+1;
if(maxx<y)
maxx=y;
}
for(i=0;i<maxx;i++)
{
while(i!=0&&color[i]&&color[i]==color[i-1])
i++;
if(color[i])
cnt[color[i]-1]++;
}
for(int i=0;i<=N;i++)
if(cnt[i])
{
printf("%d %d\n",i,cnt[i]);
}
printf("\n");
}
return 0;
}
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原文地址：http://www.cnblogs.com/a972290869/p/4254704.html
