poj 2528(线段树)

poj 2528
线段树离散化区间后区间染色，注意离散化问题

#include<cstdio>  
#include<algorithm>  
#include<cstring>  
#include<vector>
#define ms(i,j) memset(i,j, sizeof i);
using namespace std;
const int MAXN = 10000 + 5; 
const int LEFT = 0, RIGHT = 1;
struct poi
{
    int a, from, f;
    //a为坐标，from为该点来源，f为是否左右结点
    bool operator <(const poi &b) const
    {
        return a<b.a;
    } 
}p[MAXN*2];
struct ls
{
    int ch[2];//离散化后的线段 
}seg[MAXN];
int n;

#define lc o*2
#define rc o*2+1
#define M (l+r)/2
struct st
{
    int col[MAXN*4*4];//为-1则无颜色 
    void pushdown(int o)
    {
        if (col[o]!=-1)
        {
            col[lc] = col[rc] = col[o];
            col[o] = -1;
        } 
    }
    void build(int o, int l ,int r)
    {
        col[o] = -1;
        if (l==r) return ;
        build(lc, l, M);
        build(rc, M+1, r);
    }
    void update(int o, int l, int r, int x, int y, int c)
    {
        if (x<=l&&r<=y)
        {
            col[o] = c;
            return ;
        }
        pushdown(o);
        if (x<=M) update(lc, l, M, x, y, c);
        if (M<y) update(rc, M+1, r, x, y, c);
    }
    bool used[MAXN*2*4];
    int query(int o, int l, int r, int x, int y)
    {
        if (col[o]!=-1)
        {
            if (!used[col[o]]) 
            {
                used[col[o]] = true;
                return 1; 
            }else {
                return 0;
            }
        }
        int ans = 0;
        if (x<=M) ans += query(lc, l, M, x, y);
        if (M<y) ans += query(rc, M+1, r, x, y);
        return ans;
    }
}tree;
int main()  
{  
    int kase;
    scanf("%d", &kase);
    while (kase--)
    {
        scanf("%d", &n);
        for (int i=0;i<n;i++)
        {
            int x,y;
            scanf("%d%d", &x, &y);
            p[i*2].a = x, p[i*2].from = i, p[i*2].f = LEFT;
            p[i*2+1].a = y, p[i*2+1].from = i, p[i*2+1].f = RIGHT;
        } 
        sort(p, p+n*2);

        //处理第一个 
        int last = p[0].a; 
        int now = 1;
        seg[p[0].from].ch[p[0].f] = now;

        for (int i=1;i<2*n;i++)
        {
            if (p[i].a!=p[i-1].a)now++;
            if (p[i].a-last>1) now++; //离散化要在相距大于1的两个数之间加一个数！ 
            seg[p[i].from].ch[p[i].f] = now;
            last = p[i].a; 
        }
        tree.build(1,1,now);
        for (int i=0;i<n;i++) tree.update(1,1,now,seg[i].ch[0], seg[i].ch[1], i);
        ms(tree.used, false);
        printf("%d\n", tree.query(1,1,now,1,now));
    }
    return 0;  
}