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POJ 2777 Count Color 线段树

2011-08-07 13:28 302 查看
 题意:对长木板进行涂色,有两种操作, 一种插入,一种统计,输出一定范围内的颜色种数。

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
#define L(u) (u<<1)
#define R(u) (u<<1|1)
#define N 150000
struct TreeNode { int l,r,c;};
TreeNode node[N*3];
int kind, mark[31];

void build ( int u, int l, int r )
{
node[u].l = l;
node[u].r = r;
node[u].c = 1;
if ( l == r ) return;
int mid = ( l + r ) >> 1;
build ( L(u), l, mid );
build ( R(u), mid + 1, r );
}

void update ( int u, int l, int r, int c )
{
if ( node[u].c == c ) return;
if ( node[u].l == l && node[u].r == r )
{
node[u].c = c; return;
}

if ( node[u].c != -1 )
{
node[L(u)].c = node[R(u)].c = node[u].c;
node[u].c = -1;
}
int mid = (node[u].l + node[u].r) >> 1;
if ( r <= mid ) update (L(u),l,r,c);
else if ( l > mid ) update (R(u),l,r,c);
else update (L(u),l,mid,c), update(R(u),mid+1,r,c);
}

void query ( int u, int l, int r )
{
if ( node[u].c != -1 )
{
if( !mark[node[u].c] )
{
mark[node[u].c] = 1;
kind++;
}
return;
}
int mid = (node[u].l + node[u].r) >> 1;
if ( r <= mid ) query (L(u),l,r);
else if ( l > mid ) query (R(u),l,r);
else query(L(u),l,mid), query(R(u),mid+1,r);
}

int main()
{
int L, T, O, a, b, c;
char oper[10];
scanf("%d%d%d", &L, &T, &O);
build ( 1, 1, L );
while ( O-- )
{
scanf("%s %d %d",oper, &a, &b);
if ( a > b ) swap(a,b);
if ( oper[0] == 'P' )
{
kind = 0;
memset(mark,0,sizeof(mark));
query ( 1, a, b );
printf("%d\n", kind );
}
else
{
scanf("%d",&c);
update ( 1, a, b, c );
}
}
return 0;
}


 
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标签:  query build c struct