Dynamic Rankings（动态第k大+树套树）

题目链接：

　　http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1112

题目：

思路：

　　树套树板子题。

代码实现如下：

 

  1 #include <set>
  2 #include <map>
  3 #include <deque>
  4 #include <queue>
  5 #include <stack>
  6 #include <cmath>
  7 #include <ctime>
  8 #include <bitset>
  9 #include <cstdio>
 10 #include <string>
 11 #include <vector>
 12 #include <cstdlib>
 13 #include <cstring>
 14 #include <iostream>
 15 #include <algorithm>
 16 using namespace std;
 17 
 18 typedef long long LL;
 19 typedef pair<LL, LL> pLL;
 20 typedef pair<LL, int> pLi;
 21 typedef pair<int, LL> pil;;
 22 typedef pair<int, int> pii;
 23 typedef unsigned long long uLL;
 24 
 25 #define lson rt<<1
 26 #define rson rt<<1|1
 27 #define lowbit(x) x&(-x)
 28 #define name2str(name) (#name)
 29 #define bug printf("*********\n")
 30 #define debug(x) cout<<#x"=["<<x<<"]" <<endl
 31 #define FIN freopen("D://code//in.txt","r",stdin)
 32 #define IO ios::sync_with_stdio(false),cin.tie(0)
 33 
 34 const double eps = 1e-8;
 35 const int mod = 1000000007;
 36 const int maxn = 5e4 + 7;
 37 const double pi = acos(-1);
 38 const int inf = 0x3f3f3f3f;
 39 const LL INF = 0x3f3f3f3f3f3f3f3fLL;
 40 
 41 char op[5];
 42 int n, q, x, y, tot1, tot2, cnt;
 43 int a[maxn], q1[maxn], q2[maxn], root1[maxn], root2[maxn];
 44 vector<int> v;
 45 
 46 struct que {
 47     int op, l, r, x;
 48 }ask[maxn];
 49 
 50 struct node {
 51     int l, r, sum;
 52 }tree[maxn*40];
 53 
 54 int getid(int x) {
 55     return lower_bound(v.begin(), v.end(), x) - v.begin() + 1;
 56 }
 57 
 58 void update(int l, int r, int& x, int y, int pos, int val) {
 59     tree[++cnt] = tree[y], tree[cnt].sum += val, x = cnt;
 60     if(l == r) return;
 61     int mid = (l + r) >> 1;
 62     if(pos <= mid) update(l, mid, tree[x].l, tree[y].l, pos, val);
 63     else update(mid + 1, r, tree[x].r, tree[y].r, pos, val);
 64 }
 65 
 66 int query(int l, int r, int x, int y, int k) {
 67     if(l == r) return l;
 68     int cnt = tree[tree[y].l].sum - tree[tree[x].l].sum;
 69     for(int i = 0; i < tot1; ++i) {
 70         cnt -= tree[tree[q1[i]].l].sum;
 71     }
 72     for(int i = 0; i < tot2; ++i) {
 73         cnt += tree[tree[q2[i]].l].sum;
 74     }
 75     int mid = (l + r) >> 1;
 76     if(cnt >= k) {
 77         for(int i = 0; i < tot1; ++i) {
 78             q1[i] = tree[q1[i]].l;
 79         }
 80         for(int i = 0; i < tot2; ++i) {
 81             q2[i] = tree[q2[i]].l;
 82         }
 83         return query(l, mid, tree[x].l, tree[y].l, k);
 84     } else {
 85         for(int i = 0; i < tot1; ++i) {
 86             q1[i] = tree[q1[i]].r;
 87         }
 88         for(int i = 0; i < tot2; ++i) {
 89             q2[i] = tree[q2[i]].r;
 90         }
 91         return query(mid + 1, r, tree[x].r, tree[y].r, k - cnt);
 92     }
 93 }
 94 
 95 int main(){
 96 #ifndef ONLINE_JUDGE
 97     FIN;
 98 #endif
 99     int t;
100     scanf("%d", &t);
101     while(t--) {
102         scanf("%d%d", &n, &q);
103         cnt = tot1 = tot2 = 0;
104         v.clear();
105         memset(root2, 0, sizeof(root2));
106         for(int i = 1; i <= n; ++i) {
107             scanf("%d", &a[i]);
108             v.push_back(a[i]);
109         }
110         for(int i = 1; i <= q; ++i) {
111             scanf("%s", op);
112             if(op[0] == 'C') ask[i].op = 1;
113             else ask[i].op = 2;
114             if(ask[i].op == 1) {
115                 scanf("%d%d", &ask[i].l, &ask[i].x);
116                 v.push_back(ask[i].x);
117             } else {
118                 scanf("%d%d%d", &ask[i].l, &ask[i].r, &ask[i].x);
119             }
120         }
121         sort(v.begin(), v.end());
122         v.erase(unique(v.begin(), v.end()), v.end());
123         int sz = v.size();
124         for(int i = 1; i <= n; ++i) {
125             a[i] = getid(a[i]);
126             update(1, sz, root1[i], root1[i-1], a[i], 1);
127         }
128         for(int i = 1; i <= q; ++i) {
129             int op = ask[i].op;
130             if(op == 1) {
131                 int pos = ask[i].l, x = ask[i].x, y = a[pos];
132                 x = getid(x);
133                 a[pos] = x;
134                 while(pos <= n) {
135                     update(1, sz, root2[pos], root2[pos], y, -1);
136                     update(1, sz, root2[pos], root2[pos], x, 1);
137                     pos += lowbit(pos);
138                 }
139             } else {
140                 int l = ask[i].l, r = ask[i].r, k = ask[i].x;
141                 int x = l - 1;
142                 tot1 = tot2 = 0;
143                 while(x) {
144                     q1[tot1++] = root2[x];
145                     x -= lowbit(x);
146                 }
147                 x = r;
148                 while(x) {
149                     q2[tot2++] = root2[x];
150                     x -= lowbit(x);
151                 }
152                 printf("%d\n", v[query(1, sz, root1[l-1], root1[r], k)-1]);
153             }
154         }
155     }
156     return 0;
157 }

 

posted @ 2019-04-18 00:45  Dillonh  阅读(250)  评论(0)  编辑  收藏  举报