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题目链接: http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1610
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
在[0, 8000]涂色，后涂的会覆盖前面涂的，求每种颜色最后能分辨的的区块有多少个。
建树时，树的单位是区间，而不是点 如[0, 8000]的左右儿子应该分别是[0, 4000], [4000,8000];
询问时，需要判断相邻区间的颜色是否相同 用temp记录前一区间所记录的颜色来解决这个问题。
1 #include<iostream> 2 #include<cstring> 3 using namespace std; 4 5 int color[8010]; 6 int mark[20000]; 7 int temp; 8 9 void Build( int i, int l, int r ){ 10 mark[i] = -1;//-1表示没有涂 11 12 if( l + 1 == r ) return; 13 14 int mid = ( l + r ) >> 1; 15 Build( i << 1, l , mid ); 16 Build( ( i << 1 ) | 1, mid, r ); 17 } 18 19 void Insert( int i, int l, int r, int z, int y, int c ){ 20 if( l == r ) return; 21 if( z <= l && y >= r ){ 22 mark[i] = c; 23 return; 24 } 25 if( mark[i] == c ) return; 26 if( mark[i] >= 0 ){ 27 mark[i << 1] = mark[i]; 28 mark[( i << 1) | 1] = mark[i]; 29 mark[i] = -2;//该区间含有多个颜色 30 } 31 int mid = ( l + r ) >> 1; 32 if( y <= mid ) Insert( i << 1, l, mid, z, y, c ); 33 else if( z >= mid ) Insert( ( i << 1 ) | 1, mid, r, z, y, c ); 34 else{ 35 Insert( i << 1, l, mid, z, mid, c ); 36 Insert( ( i << 1 ) | 1, mid, r, mid, y, c ); 37 } 38 mark[i] = -2; 39 } 40 41 void Queue( int i, int l, int r){ 42 if( mark[i] == -1 ){ 43 temp = -1; 44 return; 45 } 46 if( mark[i] >= 0 ){ 47 if( temp != mark[i] ){ 48 temp = mark[i];//记录前一段的颜色 49 color[mark[i]]++; 50 } 51 return; 52 } 53 if( l + 1 != r ){ 54 int mid = ( l + r ) >> 1; 55 Queue( i << 1, l, mid ); 56 Queue( ( i << 1) | 1, mid, r ); 57 } 58 } 59 60 int main(){ 61 ios::sync_with_stdio( false ); 62 63 int n; 64 while( cin >> n ){ 65 66 Build( 1, 0, 8000 ); 67 memset( color, 0, sizeof( color ) ); 68 69 int a, b, c, max = 0; 70 for( int i = 1; i <= n; i++ ){ 71 cin >> a >> b >> c; 72 Insert( 1, 0, 8000, a, b, c ); 73 max = c > max ? c : max; 74 } 75 76 temp = -1; 77 Queue( 1, 0, 8000 ); 78 79 for( int i = 0; i <= max ; i++ ) 80 if( color[i] ) cout << i << " " << color[i] << endl; 81 82 cout << endl; 83 } 84 85 return 0; 86 }
ZOJ-1610 Count the Colors ( 线段树 )
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原文地址：http://www.cnblogs.com/hollowstory/p/5334860.html
