[BZOJ 3932][CQOI 2015]任务查询系统

将一个任务拆成两个点，排序后查询某个点上k小值的和
用可持久化线段树做一做就可以了

TAT以后写主席树一定要离散。。RE不止。。(其实是爆内存了。。)
特别注意容易写渣的地方是引用root[]数组。。

我怎么会直接就引用了下标呢QAQ。。

不长记性！

#include 
#define maxn 500010
using namespace std;

int n, m;

struct operation{
    int s, tp, v, cnt;
    bool operator<(const operation& k)const{
        return s < k.s;
    }
}q[maxn];
#define M 20000010
int lc[M], rc[M], val[M], root[maxn], size;

long long sum[M];

int build(int l, int r){
    int cur = ++ size;
    if(l == r)return cur;
    int mid = l + r >> 1;
    lc[cur] = build(l, mid);
    rc[cur] = build(mid+1, r);
    return cur;
}

long long hs[maxn];

int Insert(int rt, int l, int r, int pos, int v){
    int cur = ++ size;
    sum[cur] = sum[rt] + v * hs[pos];
    lc[cur] = lc[rt], rc[cur] = rc[rt], val[cur] = val[rt] + v;
    if(l == r)return cur;
    int mid = l + r >> 1;
    if(pos <= mid)lc[cur] = Insert(lc[rt], l, mid, pos, v);
    else rc[cur] = Insert(rc[rt], mid+1, r, pos, v);
    return cur;
}

long long ask(int rt, int l, int r, int kth){
    if(l == r)return min(val[rt], kth) * hs[l];
    int mid = l + r >> 1;
    if(val[lc[rt]] <= kth)
        return sum[lc[rt]] + ask(rc[rt], mid+1, r, kth - val[lc[rt]]);
    return ask(lc[rt], l, mid, kth);
}

int h[maxn];

int main(){
    scanf("%d%d", &m, &n);
    int s, e, p, tot = 0, mx = 0;
    for(int i = 1; i <= m; i ++){
        scanf("%d%d%d", &s, &e, &p);
        q[++ tot].s = s;
        q[tot].tp = 1;
        q[tot].v = p;
        q[++ tot].s = e + 1;
        q[tot].tp = -1;
        q[tot].v = p;
        hs[++ mx] = p;
    }
    sort(hs + 1, hs + 1 + mx);
    mx = unique(hs + 1, hs + 1 + mx) - hs - 1;
    for(int i = 1; i <= tot; i ++)
        q[i].v = lower_bound(hs + 1, hs + 1 + mx, q[i].v) - hs;
    sort(q + 1, q + 1 + tot);
    root[0] = build(1, mx);

    int j = 1, cnt = 0;
    for(int i = 1; i <= tot; i = j){
        ++ cnt;
        root[cnt] = root[cnt - 1];
        for(; j <= tot && q[i].s == q[j].s; j ++)
            root[cnt] = Insert(root[cnt], 1, mx, q[j].v, q[j].tp);
        h[cnt] = q[i].s;
    }
    long long pre = 1;
    int x, a, b, c;
    for(int i = 1; i <= n; i ++){
        scanf("%d%d%d%d", &x, &a, &b, &c);
        int k = 1 + ((long long)a * pre + b) % c;
        int y = lower_bound(h + 1, h + 1 + cnt, x) - h;
        while(y && h[y] > x)y --;
        printf("%lld\n", pre = ask(root[y], 1, mx, k));
    }
    return 0;
}