题目链接：

hdu 4866

题目大意：

给出n条直线，每次查询给出一个x，求这个x位置上方的前k个线段的高度之和。

题目分析：

  • 首先对高度进行离散化建立主席树，将每条直线对应为两个点，左端点+1,右端点-1，然后每个点建一个线段树，然后每次查询对于一个x，只需要二分查找找到时序点，然后找到线段数量为k的最左的前缀即可。

AC代码：

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
typedef long long LL;

const int MAXN=200005;
const int M=MAXN*20;
int root[MAXN];

struct LINE
{
    int x, d, id;
} line[MAXN];
int y[MAXN], ind[MAXN], to[MAXN],num_trees;
int i, j, cnt, n, m, x, p, L, R, D, X, A, B, C, k;
bool cmp(LINE x1,LINE x2)
{
    return x1.x<x2.x;
}

bool cmp2(int a,int b)
{
    return y[a]<y[b];
}

struct Tree
{
    int l,r,num;
    LL val;
}tree[5000007];

void build ( int &u , int l , int r )
{
    u = ++num_trees;
    tree[u].num = tree[u].val = 0;
    if ( l == r ) return;
    int mid = l+r>>1;
    build ( tree[u].l , l , mid );
    build ( tree[u].r , mid+1 , r );
}

void update ( int p , int &u , int l , int r , int x , int v , int z )
{
    u = ++num_trees;
    tree[u] = tree[p];
    tree[u].num += v;
    tree[u].val += z;
    if ( l == r) return;
    int mid = l+r>>1;
    if ( x > mid ) update ( tree[p].r , tree[u].r , mid+1 , r , x , v , z );
    else update ( tree[p].l , tree[u].l , l , mid , x , v , z );
}

LL query ( int u , int l , int r , int k )
{
    if ( k == 0 ) return 0;
    if ( l == r ) return tree[u].val;
    int mid = l+r>>1;
    if ( tree[tree[u].l].num >= k ) return query ( tree[u].l , l , mid , k );
    else return tree[tree[u].l].val + query ( tree[u].r , mid+1 , r , k-tree[tree[u].l].num );
}

int bsearch ( int x )
{
    int l=1,r=cnt,mid;
    while ( l != r )
    {
        mid = (l+r+1)>>1;
        if ( line[mid].x <= x ) l = mid;
        else r = mid-1;
    }
    return l;
}

int main()
{
    while(~scanf("%d%d%d%d",&n,&m,&x,&p))
    {
        cnt=0;
        num_trees=0;
        for(i=1;i<=n;i++){
            scanf("%d%d%d",&L,&R,&D);
            y[i]=D;
            ind[i]=i;
            line[++cnt].x=L, line[cnt].d=D, line[cnt].id=i;
            line[++cnt].x=R+1, line[cnt].d=-D, line[cnt].id=i;
        }
        sort(line+1,line+cnt+1,cmp);
        sort(ind+1,ind+n+1,cmp2);
        for(i=1;i<=n;i++) to[ind[i]]=i;
        build ( root[0] , 1 , n );
        for(i=1;i<=cnt;i++)
            if ( line[i].d> 0 )
                update ( root[i-1] , root[i] , 1 , n , to[line[i].id] , 1 , line[i].d );
            else
                update ( root[i-1] , root[i] , 1 , n , to[line[i].id] , -1 , line[i].d ); 
        LL  pre=1, ret;
        for(i=1;i<=m;i++){
            scanf("%d%d%d%d",&X,&A,&B,&C);
            k=(pre%C*A%C+B)%C;
            int tt = bsearch ( X );
            ret = query ( root[tt] , 1 , n , k );
            if(pre>p) ret*=2;
            pre=ret;
            printf("%I64d\n",ret);
        }
    }
    return 0;
}