HDU5649 DZY Loves Sorting 二分+线段树

上周六BC的最后一道题

感觉还是挺经典的做法的，首先因为是n的全排列，所以可以直接二分答案然后用线段树判断，好吧其实bc的题解说的已经很清楚了

这是一道良心的基础数据结构题。

我们二分$a[k]$的值，假设当前是$mid$，然后把大于$mid$的数字标为$1$，不大于$mid$的数字标为$0$。然后对所有操作做完以后检查一下$a[k]$位置上是$0$还是$1$。

因为只有两种值，所以操作还是不难做的。只要用一个线段树，支持区间求和、区间赋值即可。这样要排序一个区间时只要查询一下里面有几个$1$和几个$0$，然后把前半段赋值为$0$，后半段赋值为$1$即可（降序的话就是反过来）。

复杂度是O(m\log ^2 n)O(mlog​2​​n)的。

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#define maxn 100009#define rep(i, j, k) for(int i = j; i <= k; i++)using namespace std;int n, m, k, f[maxn], tot[6 * maxn], all[maxn * 6];int a[maxn], b[maxn], c[maxn], Ans;void insert (int x, int lx, int rx, int position, int num){if (position > rx || position < lx)return;if (lx == rx && lx == position){tot[x] = num;return;}int mid = (lx + rx) >> 1;if (position <= mid)insert (2 * x, lx, mid, position, num);elseinsert (2 * x + 1, mid + 1, rx, position, num);tot[x] = tot[2 * x] + tot[2 * x + 1];return;}void set (int x, int la, int ra, int L, int R, int num){if (R < L)return;if (L > ra || R < la)return;if (L <= la && ra <= R){all[x] = num;tot[x] = num * (ra - la + 1);return;}int mid = (la + ra) >> 1;if (all[x] != -1){all [2 * x] = all [2 * x + 1] = all [x], tot[x] = all[x] * (ra - la + 1);tot[2 * x] = all[x] * (mid - la + 1);tot[2 * x + 1] = all[x] * (ra - mid);all [x] = -1;}set (2 * x, la, mid, L, R, num);set (2 * x + 1, mid + 1, ra, L, R, num);tot[x] = tot[2 * x] + tot[2 * x + 1];return;}int ask (int x, int la, int ra, int L, int R){if (L > ra || R < la)return 0;if (L <= la && ra <= R){if (all[x] != -1)tot[x] = all[x] * (ra - la + 1), all[x] = -1;return tot[x];}int mid = (la + ra) >> 1;if (all[x] != -1){all[2 * x] = all[2 * x + 1] = all[x];tot[2 * x] = all[x] * (mid - la + 1);tot[2 * x + 1] = all[x] * (ra - mid);all[x] = -1;}int num = ask (2 * x, la, mid, L, R) + ask (2 * x + 1, mid + 1, ra, L, R);tot[x] = tot[2 * x] + tot[2 * x + 1];return num;}void work (){int l = 1, r = n;while (l <= r){memset (all, -1, sizeof (all));memset (tot, 0, sizeof (tot));int mid = (l + r) >> 1;rep (i, 1, n)insert (1, 1, n, i, f[i] >= mid ? 1 : 0);rep (i, 1, m){int now = ask (1, 1, n, b[i], c[i]);//the num of 1int other = c[i] - b[i] + 1 - now;//the num of 0if (a[i] == 0){set (1, 1, n, b[i], b[i] + other - 1, 0);set (1, 1, n, b[i] + other, c[i], 1);}else{set (1, 1, n, b[i], b[i] + now - 1, 1);set (1, 1, n, b[i] + now, c[i], 0);}}if (ask (1, 1, n, k, k) > 0)l = mid + 1, Ans = mid;elser = mid - 1;}printf ("%d\n", Ans);return ;}int main(){int Case;scanf("%d", &Case);while (Case--){scanf ("%d%d", &n, &m);rep (i, 1, n)scanf ("%d", &f[i]);rep (i, 1, m)scanf ("%d%d%d", &a[i], &b[i], &c[i]);scanf ("%d", &k);work ();}return 0;}