BZOJ K大数查询（分治）（Zjoi2013）

思路：分治答案。答案范围[-n, n]，从前往后扫描，若是插入操作且c>mid，则把线段树中区间[a, b]加一，并置为为类别1；否则置为类别0。若是询问操作，若目前线段树中区间[a, b]的和小于等于c，则置为类别1；否则置为类别0，并把c减去区间[a, b]的和。然后分治处理，其中类别0中，答案范围为[-n, mid]；类别1中，答案范围为[mid + 1, n]。按类别排序后，两个区间之间互不影响。时间复杂度为O(nlognlogn)。

  1 #include <cstdio>
  2 #include <algorithm>
  3 #include <iostream>
  4 #include <cstring>
  5 using namespace std;
  6 typedef long long LL;
  7 
  8 const int MAXN = 50010;
  9 const int MAXT = MAXN << 2;
 10 
 11 int sum[MAXT];
 12 int add[MAXT];
 13 bool clr[MAXT];
 14 
 15 #define ll (x << 1)
 16 #define rr (ll | 1)
 17 #define mid ((l + r) >> 1)
 18 void initTree() {
 19     sum[1] = add[1] = 0;
 20     clr[1] = true;
 21 }
 22 
 23 void pushdown(int x, int l, int r) {
 24     if(clr[x]) {
 25         clr[ll] = clr[rr] = true;
 26         sum[ll] = sum[rr] = add[ll] = add[rr] = 0;
 27         clr[x] = false;
 28     }
 29     if(add[x]) {
 30         sum[ll] += (mid - l + 1) * add[x];
 31         add[ll] += add[x];
 32         sum[rr] += (r - mid) * add[x];
 33         add[rr] += add[x];
 34         add[x] = 0;
 35     }
 36 }
 37 
 38 void maintain(int x) {
 39     sum[x] = sum[ll] + sum[rr];
 40 }
 41 
 42 void modify(int x, int l, int r, int a, int b) {
 43     if(a <= l && r <= b) {
 44         add[x]++;
 45         sum[x] += (r - l + 1);
 46     } else {
 47         pushdown(x, l, r);
 48         if(a <= mid) modify(ll, l, mid, a, b);
 49         if(mid < b) modify(rr, mid + 1, r, a, b);
 50         maintain(x);
 51     }
 52 }
 53 
 54 int query(int x, int l, int r, int a, int b) {
 55     if(a <= l && r <= b) {
 56         return sum[x];
 57     } else {
 58         pushdown(x, l, r);
 59         int res = 0;
 60         if(a <= mid) res += query(ll, l, mid, a, b);
 61         if(mid < b) res += query(rr, mid + 1, r, a, b);
 62         return res;
 63     }
 64 }
 65 #undef mid
 66 
 67 struct Node {
 68     int op, id, a, b, c, v;
 69     void read(int i) {
 70         id = i;
 71         scanf("%d%d%d%d", &op, &a, &b, &c);
 72     }
 73     bool operator < (const Node &rhs) const {
 74         if(v != rhs.v) return v < rhs.v;
 75         return id < rhs.id;
 76     }
 77 } p[MAXN];
 78 int ans[MAXN];
 79 int n, m;
 80 
 81 void work(int a, int b, int l, int r) {
 82     if(l > r) return ;
 83     if(a == b) {
 84         for(int i = l; i <= r; ++i)
 85             if(p[i].op == 2) ans[p[i].id] = a;
 86         return ;
 87     }
 88     initTree();
 89     int mid = a + ((b - a) >> 1), t = l - 1;
 90     for(int i = l; i <= r; ++i) {
 91         if(p[i].op == 1) {
 92             if(p[i].c > mid) modify(1, 1, n, p[i].a, p[i].b), p[i].v = 1;
 93             else p[i].v = 0;
 94         } else {
 95             int s = query(1, 1, n, p[i].a, p[i].b);
 96             if(p[i].c <= s) p[i].v = 1;
 97             else p[i].v = 0, p[i].c -= s;
 98         }
 99         t += !p[i].v;
100     }
101     sort(p + l, p + r + 1);
102     work(a, mid, l, t);
103     work(mid + 1, b, t + 1, r);
104 }
105 
106 int main() {
107     scanf("%d%d", &n, &m);
108     for(int i = 1; i <= m; ++i) p[i].read(i);
109     memset(ans + 1, 0x80, m * sizeof(int));
110     work(-n, n, 1, m);
111     for(int i = 1; i <= m; ++i)
112         if(ans[i] >= -n) printf("%d
", ans[i]);
113 }

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BZOJ K大数查询（分治）（Zjoi2013）

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