POJ 2528 Mayor's posters

这道题的离散化很巧妙，参考了胡浩大神的代码。自己先理解下

http://www.notonlysuccess.com/ 大神的博客

/*Accepted    1120K    79MS    C++    2109B    2012-07-24 17:40:47*/
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1

const int MAXN = 11111;
int li[MAXN], ri[MAXN];
int col[MAXN << 4];
int X[MAXN << 2];
int cnt;
bool hash[MAXN];

void PushDown( int rt)
{
    if( col[rt] != -1)
    {
        col[rt << 1] = col[rt << 1 | 1] = col[rt];
        col[rt] = -1;
    }
}

void update( int L, int R, int c, int l, int r, int rt)
{
    int m = (l + r) >> 1;
    if( L <= l && r <= R)
    {
        col[rt] = c;
        return;
    }
    PushDown(rt);
    if( L <= m) update( L, R, c, lson);
    if( R > m) update( L, R, c, rson);
}

void query( int l, int r, int rt)
{
    if( col[rt] != -1){
        if( !hash[col[rt]]) cnt ++;
        hash[col[rt]] = true;
        return;
    }
    if( l == r) return;
    int m = (l + r) >> 1;
    query(lson);
    query(rson);
}

int BS( int key, int n, int X[])
{
    int l = 0, r = n - 1;
    while( l <= r)
    {
        int m = (l + r) >> 1;
        if( key == X[m]) return m;
        if( X[m] < key) l = m + 1;
        else r = m - 1;
    }
    return -1;
}

int main()
{
    int T;
    int n, nn, m;
    scanf( "%d", &T);
    while( T --){
        cnt = 0;
        scanf( "%d", &n);
        nn = 0;
        for( int i = 0; i < n; i ++)
        {
            scanf( "%d%d", &li[i], &ri[i]);
            X[nn ++] = li[i];
            X[nn ++] = ri[i];
        }
        sort( X, X + nn);
        int m = 1;
        for( int i = 1; i < nn; i ++)
            if( X[i] != X[i - 1]) X[m ++] = X[i];
        for( int i = m - 1; i > 0; i --)
            if( X[i] != X[i - 1] + 1) X[m ++] = X[i - 1] + 1;
        sort(X, X + m);
        memset( col, -1, sizeof col);
        memset( hash, false, sizeof hash);
        for( int i = 0; i < n; i ++)
        {
            int l = BS( li[i], m, X);
            int r = BS( ri[i], m, X);
            update( l, r, i, 0, m, 1);
        }
        query( 0, m, 1);
        printf( "%d\n", cnt);
    }
    return 0;
}