ZOJ1610
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
#include
#define R(x) (x << 1 | 1)
using namespace std;
const int MAX = 8010; struct Tnode{ int col; int l,r;}; Tnode node[MAX*3]; int col[MAX],len[MAX];; void init() {
memset(node,0,sizeof(node)); memset(col,-1,sizeof(col)); }
void Build(int t,int l,int r) {
node[t].l = l; node[t].r = r; node[t].col = -1; if( l == r - 1 ) return ; int mid = ( l + r ) >> 1; Build(L(t), l, mid); Build(R(t), mid, r); }
void Updata(int t,int l,int r,int col) {
if( node[t].l >= l && node[t].r <= r ) {
node[t].col = col; return ; }
if( node[t].col == col ) return ; if( node[t].col >= 0 ) {
node[R(t)].col = node[t].col; node[L(t)].col = node[t].col; node[t].col = -1; }
int mid = (node[t].l + node[t].r) >> 1; if( l >= mid )
Updata(R(t),l,r,col); else
if( r <= mid )
Updata(L(t),l,r,col); else {
Updata(L(t),l,mid,col); Updata(R(t),mid,r,col);
} }
void Compute(int t,int l,int r) {
if( node[t].col >= 0 ) {
for(int i=l; i if( node[t].l == node[t].r - 1 ) return ;// 如果父亲节点已经是这种颜色了，就没必要再染色了 int mid = (node[t].l + node[t].r) >> 1; if( l >= mid ) Compute(R(t),l,r); else if( r <= mid ) Compute(L(t),l,r); else { Compute(L(t),l,mid); Compute(R(t),mid,r); } }