BZOJ3932: [CQOI2015]任务查询系统

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题目大意：在[s,e]（闭区间）上加入一个数字k（就相当于在s时刻放入，e+1时刻取出）。支持询问在t时刻时的第k大数。

想一想就可以知道主席树能水过这道题。

/**************************************************************    Problem: 3932    User: geng4512    Language: C++    Result: Accepted    Time:3808 ms    Memory:192900 kb****************************************************************/#include<cstdio>#include<vector>#include<algorithm>#define MAXN 100005#define LL long longusing namespace std;struct node {    int lc, rc, cnt;    long long sum;} t[MAXN*80];vector<int> L[MAXN], R[MAXN];int n, m, rt[MAXN], cnt, tag, mark[MAXN*80];char c;inline void GET(int &n) {    n = 0;    do c = getchar(); while('0' > c || c > '9');    do n = n*10+c-'0', c = getchar(); while('0' <= c && c <= '9');}inline void Build(int &rt, int x) {    t[rt = ++ cnt] = t[x];    mark[rt] = tag;}void Insert(int &rt, int p, int v, int l, int r, char f) {    if(mark[p] != tag) Build(rt, p);    t[rt].cnt += f;    t[rt].sum += v*f;    if(l >= r) return;    int mid = (l + r) >> 1;    if(v <= mid) Insert(t[rt].lc, t[p].lc, v, l, mid, f);    else Insert(t[rt].rc, t[p].rc, v, mid+1, r, f);}long long Query(int x, int l, int r, int k) {    if(l >= r) return 1LL * k * l;    int mid = (l + r) >> 1;    if(t[t[x].lc].cnt >= k) return Query(t[x].lc, l, mid, k);    return Query(t[x].rc, mid+1, r, k - t[t[x].lc].cnt) + t[t[x].lc].sum;}int main() {    int s, e, v, x, mn=MAXN, mx=-1;    long long p = 1;    scanf("%d%d", &n, &m);    for(int i = 1; i <= n; ++ i) {        GET(s); GET(e); GET(v);        L[s].push_back(v); R[e].push_back(v);        mn = min(mn, s); mx = max(mx, e);    }    for(int i = mn; i <= mx; ++ i) {        Build(rt[i], rt[i-1]); ++tag;        for(int j = 0; j < (int)L[i].size(); ++ j)            Insert(rt[i], rt[i], L[i][j], 1, 1e7, 1);        for(int j = 0; j < (int) R[i-1].size(); ++ j)            Insert(rt[i], rt[i], R[i-1][j], 1, 1e7, -1);    }    for(int i = 1; i <= m; ++ i) {        GET(x); GET(s); GET(e); GET(v);        v = 1 + (p*s+e)%v;        if(v >= t[rt[x]].cnt) p = t[rt[x]].sum;        else p = Query(rt[x], 1, 1e7, v);        printf("%lld\n", p);    }    return 0;}

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