3110: [Zjoi2013]K大数查询

3110: [Zjoi2013]K大数查询

https://lydsy.com/JudgeOnline/problem.php?id=3110

分析：

　　整体二分+线段树。

　　两种操作：区间加入一个数，区间询问第k大值。

　　如果只有一种操作，我们可以二分答案x，然后把大于x的都加入到线段树中去（区间[l,r]整体+1）,然后查询这次询问的区间有多少数（区间[l,r]求和）。

　　多种操作的话整体二分就行了，注意到有时间顺序，所以可以按照时间顺序加入和查询。

　　注意一下要开longlong。

代码：

 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 #include<iostream>
 #include<cmath>
 #include<cctype>
 #include<set>
 #include<queue>
 #include<vector>
 #include<map>
 #define Root 1, n, 1
 #define lson l, mid, rt << 1
 #define rson mid + 1, r, rt << 1 | 1
 using namespace std;
 typedef long long LL;

 inline int read() {
     int x=,f=;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-;
     for(;isdigit(ch);ch=getchar())x=x*+ch-'';return x*f;
 }

 const int N = ;

 int ans[N], n, m;
 struct Que{
     int ty, l, r, v, id;
     bool operator < (const Que &A) const {
         return id < A.id;
     }
 }A[N], tl[N], tr[N];
 struct SegmentTree{
     LL sum[N << ], tag[N << ];
     void pushup(int rt) { sum[rt] = sum[rt << ] + sum[rt <<  | ]; }
     void pushdown(int rt,int len) {
         sum[rt << ] += 1ll * (len - len / ) * tag[rt];
         sum[rt <<  | ] += 1ll * (len / ) * tag[rt];
         tag[rt << ] += tag[rt];
         tag[rt <<  | ] += tag[rt];
         tag[rt] = ;
     }
     void update(int l,int r,int rt,int L,int R,int v) {
         if (L <= l && r <= R) {
             tag[rt] += v, sum[rt] += (r - l + ) * v; return ;
         }
         if (tag[rt]) pushdown(rt, r - l + );
         int mid = (l + r) >> ;
         if (L <= mid) update(lson, L, R, v);
         if (R > mid) update(rson, L, R, v);
         pushup(rt);
     }
     LL query(int l,int r,int rt,int L,int R) {
         if (L <= l && r <= R) return sum[rt];
         if (tag[rt]) pushdown(rt, r - l + );
         int mid = (l + r) >> ; LL res = ;
         if (L <= mid) res += query(lson, L, R);
         if (R > mid) res += query(rson, L, R);
         return res;
     }
 }T;

 void solve(int l,int r,int Head,int Tail) {
     if (Head > Tail) return ;
     if (l == r) {
         for (int i = Head; i <= Tail; ++i)
             if (A[i].ty == ) ans[A[i].id] = l;
         return ;
     }
     int mid = (l + r + ) >> , cl = , cr = ;
     for (int i = Head; i <= Tail; ++i) {
         if (A[i].ty == ) {
             if (A[i].v >= mid) T.update(Root, A[i].l, A[i].r, ), tr[++cr] = A[i];
             else tl[++cl] = A[i];
         }
         else {
             LL t = T.query(Root, A[i].l, A[i].r);
             if (t >= A[i].v) tr[++cr] = A[i];
             else A[i].v -= t, tl[++cl] = A[i];
         }
     }
     for (int i = Head; i <= Tail; ++i) if (A[i].ty ==  && A[i].v >= mid) T.update(Root, A[i].l, A[i].r, -);
     for (int i = ; i <= cl; ++i) A[i + Head - ] = tl[i];
     for (int i = ; i <= cr; ++i) A[i + Head + cl - ] = tr[i];
     solve(l, mid - , Head, Head + cl - );
     solve(mid, r, Head + cl, Tail);
 }
 int main() {
     n = read(), m = read(); int Q = ;
     for (int i = ; i <= m; ++i) {
         A[i].ty = read(), A[i].l = read(), A[i].r = read(), A[i].v = read(), A[i].id = ;
         if (A[i].ty == ) A[i].id = ++Q;
     }
     solve(, n, , m);
     for (int i = ; i <= Q; ++i) printf("%d\n",ans[i]);
     return ;
 }