题意：给定一些操作，不断更新一些区间的颜色，求最后间断每种颜色的个数，直接线段树处理，每次更新，加上lazy标记，注意如何处理区间的端点是这题的关键;

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<vector>
#include<map>
#include<set>
#include<cstring>
#define INF 0x3f3f3f3f
#define N 8005
using namespace std;
typedef struct node{int x;int y;int date;
}node;
node a[N*4];
int b[N*4];
int p[N*4];
void built(int root,int first,int end){if(first==end){a[root].date=-1;a[root].x=first;a[root].y=end;return ;}int mid=(first+end)/2;built(root*2,first,mid);built(root*2+1,mid+1,end);a[root].x=a[root*2].x;a[root].y=a[root*2+1].y;a[root].date=-1;
}
void U(int root,int first,int end,int l,int r,int e){if(l<=first&&end<=r){a[root].date=e;return ;}if(a[root].date!=-1){a[root*2].date=a[root].date;a[root*2+1].date=a[root].date;a[root].date=-1;}int mid=(first+end)/2;if(l<=mid) U(root*2,first,mid,l,r,e);if(r>mid)  U(root*2+1,mid+1,end,l,r,e);
}
void Q(int root,int first,int end){if(first==end){b[first]=a[root].date;return ;}if(a[root].date!=-1){a[root*2].date=a[root].date;a[root*2+1].date=a[root].date;a[root].date=-1;}int mid=(first+end)/2;Q(root*2,first,mid);Q(root*2+1,mid+1,end);
}
int main(){int m;while(scanf("%d",&m)==1){int sum=-INF;int ans=-INF;built(1,0,N);memset(b,-1,sizeof(b));int c,d,e;for(int i=1;i<=m;i++){scanf("%d %d %d",&c,&d,&e);sum=max(sum,d-1);ans=max(ans,e);U(1,0,N,c,d-1,e);}Q(1,0,N);memset(p,0,sizeof(p));for(int i=0;i<=sum;i++){if(i==0){if(b[i]!=-1){p[b[i]]++;}}else{if(b[i]!=b[i-1]&&b[i]!=-1){p[b[i]]++;}}}for(int i=0;i<=ans;i++){if(p[i]!=0){printf("%d %d\n",i,p[i]);}}printf("\n");}return 0;
}