ZOJ 2112-Dynamic Rankings （树状数组+主席树）

题意：给你n个数，m次操作：

Q x y z表示查询区间[x,y]中第z小的数是多少；

C x y  表示将第x个位置上的数替换成y；

题解：和http://blog.csdn.net/haut_ykc/article/details/77678882 一样，只是多了修改操作；

要知道每次修改某一个位置上的数都会影响后边的所有线段树，假如暴力更新的话必然会超时。

我们可以考虑用树状数组，因此可以考虑把主席树分成两部分，一部分表示原序列、另一部分表示修改过的即可。

#include<set>#include<map>       #include<stack>              #include<queue>              #include<vector>      #include<string>    #include<math.h>     #include<stdio.h>              #include<iostream>              #include<string.h>              #include<stdlib.h>      #include<algorithm>     #include<functional>      using namespace std;    typedef long long ll;#define inf  1000000000         #define mod 1000000007               #define maxn  2600005   #define PI 3.1415926  #define lowbit(x) (x&-x)                          #define eps 1e-9         int a[60005], b[60005], rt[60005], num, f[60005], used[60005];int cnt, sum[maxn], ls[maxn], rs[maxn], n, q[10005][4];void build(int &id, int l, int r){id = ++cnt;sum[id]=0;if (l == r)return;int mid = (l + r) / 2;build(ls[id], l, mid);build(rs[id], mid + 1, r);}void update(int &id, int l, int r, int last, int val, int k){id = ++cnt;ls[id] = ls[last];rs[id] = rs[last];sum[id] = sum[last] + k;if (l == r)return;int mid = (l + r) / 2;if (val <= mid)update(ls[id], l, mid, ls[last], val, k);elseupdate(rs[id], mid + 1, r, rs[last], val, k);}      int sm(int x){int ans = 0;while (x){ans += sum[ls[used[x]]];x -= lowbit(x);}   return ans; }int query(int st, int ed, int val){int i, les, res;int l = 1, r = num;les = rt[st];res = rt[ed];for (i = st;i > 0;i -= lowbit(i))    used[i] = f[i];for (i = ed;i > 0;i -= lowbit(i))used[i] = f[i];while (l < r)      {int mid = (l + r) / 2;int tmp = sm(ed) - sm(st) + sum[ls[res]] - sum[ls[les]];if (val <= tmp){r = mid;for (i = st;i > 0;i -= lowbit(i))used[i] = ls[used[i]];for (i = ed;i > 0;i -= lowbit(i))used[i] = ls[used[i]];les = ls[les];res = ls[res];}else{l = mid + 1;val -= tmp;for (i = st;i > 0;i -= lowbit(i))used[i] = rs[used[i]];for (i = ed;i > 0;i -= lowbit(i))used[i] = rs[used[i]];les = rs[les];res = rs[res];   }}return l;}void add(int x, int v, int k){while (x <= n){int tmp = f[x];update(f[x], 1, num, tmp, v, k);x += lowbit(x);}}int main(void){char c;int T, m, i, j;scanf("%d", &T);while (T--){cnt = 0;memset(sum, 0, sizeof(sum));scanf("%d%d", &n, &m);for (i = 1;i <= n;i++)scanf("%d", &a[i]), b[i] = a[i];num = n;for (i = 1;i <= m;i++){scanf(" %c", &c);if (c == 'Q'){scanf("%d%d%d", &q[i][1], &q[i][2], &q[i][3]);q[i][0] = 1;   }else {scanf("%d%d", &q[i][1], &q[i][2]);b[++num] = q[i][2];q[i][0] = 0;}}sort(b + 1, b + num + 1);num = unique(b + 1, b + num + 1) - b - 1;    for (i = 1;i <= num;i++)a[i] = lower_bound(b + 1, b + num + 1, a[i]) - b;build(rt[0], 1, num);for (i = 1;i <= n;i++)update(rt[i], 1, num, rt[i - 1], a[i], 1);for (i = 1;i <= n;i++)f[i] = rt[0];for (i = 1;i <= m;i++){if (q[i][0] == 1){int ans = query(q[i][1]-1, q[i][2], q[i][3]);printf("%d\n", b[ans]);}else{int tmp;add(q[i][1], a[q[i][1]], -1);tmp = lower_bound(b + 1, b + num + 1, q[i][2]) - b;add(q[i][1], tmp, 1);a[q[i][1]] = tmp;}    }}return 0;}/*10005 51 2 3 4 5Q 2 3 1Q 1 5 5C 3 1Q 2 3 1Q 2 3 25 4-1 -2 -3 -4 -5Q 1 3 2Q 1 3 1C 2 1Q 1 5 4*/