hdu 4864(2) 线段树

对task和machine的yi由小到大进行排序，然后对machine来跟task配对。当machine[].yi >= task[].yi时，就更新线段树，在1-1440上做线段树,线段树存的是task[].xi，同时用用优先队列保存task[].yi；当machine[].yi < task[].yi时，就查找 1到machine[].xi最大的值。如果存在最大值的话，把优先队列里的task[].yi取出来。。这样一个machine就匹配到了一个最优的任务。还是看代码好好意会吧，细节挺多的

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<string>
#include<queue>
#include<cmath>
#include<map>

using namespace std;

#define mnx 100050
#define ll long long
#define inf 0x3f3f3f3f
#define lson l, m, rt << 1
#define rson m+1, r, rt << 1 | 1

priority_queue<int> q[1500];
int cnt[1500], sum[8000];
struct s{
    int xi, yi;
    bool operator < ( const s & b ) const{
        return yi < b.yi;
    }
}a[mnx], b[mnx];
void pushup( int rt ){
    sum[rt] = max( sum[rt<<1], sum[rt<<1|1] );
}
void build( int l, int r, int rt ){
    if( l == r ){
        sum[rt] = -1;
        return ;
    }
    int m = ( l + r ) >> 1;
    build( lson ), build( rson );
    pushup( rt );
}
int find( int L, int R, int l, int r, int rt ){
    if( L <= l && R >= r ){
        return sum[rt];
    }
    int m = ( l + r ) >> 1;
    int ret = -1;
    if( L <= m ) ret = max( ret, find( L, R, lson ) );
    if( R > m ) ret = max( ret, find( L, R, rson ) );
    return ret;
    
}
void update( int u, int v, int l, int r, int rt ){
    if( l == r ){
        sum[rt] = v;
        return ;
    }
    int m = ( l + r ) >> 1;
    if( u <= m ) update( u, v, lson );
    else update( u, v, rson );
    pushup( rt );
}
int main(){
    int n, m;
    const int nn = 1450;
    while( scanf( "%d%d", &n, &m ) != EOF ){
        for( int i = 0; i < nn; i++ ){
            while( !q[i].empty() ) q[i].pop();
        }
        memset( cnt, 0, sizeof(cnt) );
        for( int i = 0; i < n; i++ ){
            scanf( "%d%d", &a[i].xi, &a[i].yi );
        }
        for( int i = 0; i < m; i++ ){
            scanf( "%d%d", &b[i].xi, &b[i].yi );
        }
        sort( a, a+n );
        sort( b, b+m );
        build( 1, nn, 1 );
        int j = 0, ans1 = 0; ll ans2 = 0;
        for( int i = 0; i < n; i++ ){
            while( j < m && a[i].yi >= b[j].yi ){
                if( !cnt[b[j].xi] ){
                    update( b[j].xi, b[j].xi, 1, nn, 1 );
                }
                cnt[b[j].xi]++;
                q[b[j].xi].push( b[j].yi );
                j++;
            }
            int t = find( 1, a[i].xi, 1, nn, 1 );
            if( t == - 1 ) continue;
            cnt[t]--;
            if( cnt[t] == 0 ) update( t, -1, 1, nn, 1 );
            int tt = q[t].top();  q[t].pop();
            ans1++;
            ans2 += t * 500 + tt * 2;
        }
        printf( "%d %I64d
", ans1, ans2 );
    }
    return 0;
}