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
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
ZOJ挂了，我还没交没不知道对不对。应该没问题，按照kuangbin大神的模板改的。这个题又是区间更新，不过看数据范围不用离散化。
因为每个节点都代表区间的端点，kuangbin大神的处理方法在建树的时候与普通线段树模板做了微小的改动。
代码如下：
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 #define M 8010 5 struct segTree 6 { 7 int l,r,col; 8 }tree[M<<2]; 9 int color[M],temp,n; 10 void buildTree (int now,int l,int r) 11 { 12 tree[now].l=l,tree[now].r=r; 13 tree[now].col=-1;//-1代表没有颜色 14 if (l+1==r) 15 return ; 16 int m=((l+r)>>1); 17 buildTree(now<<1,l,m); 18 buildTree(now<<1|1,m,r);//因为节点代表端点，二分建树时是[l,m],[m,r],而不是[m+1,r]。 19 } 20 void upDate (int now,int l,int r,int c) 21 { 22 if (l==r||l>r) 23 return ; 24 if (tree[now].col==c) 25 return ; 26 if (l<=tree[now].l&&tree[now].r<=r) 27 { 28 tree[now].col=c; 29 return ; 30 } 31 if (tree[now].col>=0) 32 { 33 tree[now<<1].col=tree[now].col; 34 tree[now<<1|1].col=tree[now].col; 35 tree[now].col=-2; 36 } 37 int mid=(tree[now].l+tree[now].r)>>1; 38 if (r<=mid) 39 upDate(now<<1,l,r,c); 40 else if (l>=mid) 41 upDate(now<<1|1,l,r,c); 42 else 43 { 44 upDate(now<<1,l,mid,c); 45 upDate(now<<1|1,mid,r,c); 46 } 47 tree[now].col=-2;//-2代表有多种颜色 48 } 49 void Count (int now) 50 { 51 if (tree[now].col==-1) 52 { 53 temp=-1;//temp是前一段的颜色 54 return ; 55 } 56 if (tree[now].col!=-2) 57 { 58 if (tree[now].col!=temp) 59 { 60 color[tree[now].col]++; 61 temp=tree[now].col; 62 } 63 return ; 64 } 65 if (tree[now].l+1!=tree[now].r) 66 { 67 Count(now<<1); 68 Count(now<<1|1); 69 } 70 } 71 int main() 72 { 73 //freopen("de.txt","r",stdin); 74 while (~scanf("%d",&n)) 75 { 76 int x,y,z,maxn=0; 77 buildTree(1,0,8000); 78 for (int i=0;i<n;++i) 79 { 80 scanf("%d%d%d",&x,&y,&z); 81 upDate(1,x,y,z); 82 maxn=max(maxn,z); 83 } 84 temp=-1; 85 memset(color,0,sizeof color); 86 Count(1); 87 for (int i=0;i<=maxn;++i) 88 { 89 if (color[i]) 90 printf("%d %d ",i,color[i]); 91 } 92 printf(" "); 93 } 94 return 0; 95 }