[CQOI 2015] 任务查询系统

[题目链接]

https://www.lydsy.com/JudgeOnline/problem.php?id=3932

[算法]

首先 ， 我们可以将(Si , Ei , Pi)转化为在Si处加入Pi ， 在(Ei + 1)出删除Pi

建立可持久化线段树 ， 维护每秒出现任务的个数和优先级的和 ， 即可

时间复杂度 : O(NlogN)

[代码]

#include<bits/stdc++.h>
using namespace std;
#define MAXN 200010
typedef long long LL;

int n , m , idx , version;
int S[MAXN] , E[MAXN] , P[MAXN] , root[MAXN * ] , cnt[MAXN * ] , tmp[MAXN * ] , lson[MAXN * ] , rson[MAXN * ] , rt[MAXN * ];
LL sum[MAXN * ];
multiset< int > ins[MAXN] , del[MAXN];

template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }
template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); }
template <typename T> inline void read(T &x)
{
    T f = ; x = ;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
    for (; isdigit(c); c = getchar()) x = (x << ) + (x << ) + c - '';
    x *= f;
}
inline void build(int &k , int l , int r)
{
        k = ++idx;
        cnt[k] = sum[k] = ;
        if (l == r) return;
        int mid = (l + r) >> ;
        build(lson[k] , l , mid);
        build(rson[k] , mid +  , r);
}
inline void modify(int &k , int old , int l , int r , int x , int value)
{
        k = ++idx;
        lson[k] = lson[old] , rson[k] = rson[old];
        cnt[k] = cnt[old] + value;
        sum[k] = sum[old] + 1LL * tmp[x] * value;
        if (l == r) return;
        int mid = (l + r) >> ;
        if (mid >= x) modify(lson[k] , lson[k] , l , mid , x , value);
        else modify(rson[k] , rson[k] , mid +  , r , x , value);
}
inline LL query(int &k , int l , int r , int x)
{
    if (l == r) return 1LL * tmp[l] * min(x , cnt[k]);
    int mid = (l + r)    >> ;
    if (cnt[lson[k]] >= x) return query(lson[k] , l , mid , x);
    else return sum[lson[k]] + query(rson[k] , mid +  , r , x - cnt[lson[k]]);
}

int main()
{

        read(n); read(m);
        for (int i = ; i <= n; i++)
        {
              read(S[i]);
                read(E[i]);
              read(P[i]);
              tmp[i] = P[i];
        }
        sort(tmp +  , tmp + n + );
        int len = unique(tmp +   , tmp + n + ) - tmp - ;
        for (int i = ; i <= n; i++)
        {
                P[i] = lower_bound(tmp +   , tmp + len +  , P[i]) - tmp;
                ins[S[i]].insert(P[i]);
                del[E[i] + ].insert(P[i]);
        }
        build(root[version = ] ,  , len);
        for (int i = ; i <= n; i++)
        {
                for (multiset< int > :: iterator it = ins[i].begin(); it != ins[i].end(); it++)
                {
                        int t = (*it);
                        modify(root[++version] , root[version] ,  , n , t , );
                }
                for (multiset< int > :: iterator it = del[i].begin(); it != del[i].end(); it++)
                {
                        int t = (*it);
                        modify(root[++version] , root[version] ,  , n , t , -);
                }
                rt[i] = version;
        }
        LL pre = ;
        for (int i = ; i <= m; i++)
        {
            LL xi , ai , bi , ci;
            read(xi); read(ai); read(bi); read(ci);
            int ki = (1LL * ai * (pre % ci) % ci + bi % ci) % ci + ;
            printf("%lld\n" , pre = query(root[rt[xi]] ,  , n , ki));
        }

        return ;

}