# 【BZOJ 1901】【ZJU 2112】Dynamic Rankings

http://www.lydsy.com/JudgeOnline/problem.php?id=1901

cdq是对时间分治，整体二分是对答案分治。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int N = 10003;

struct node {int x, y, k, op, id;} B[N << 2], q1[N << 2], q2[N << 2];
int tot = 0, a[N], n, m, bits[N << 1], top = 0, H[N << 1], cnt = 0;
int ans[N], id[N << 2], anstot = 0, que[N << 1], q1len, q2len;

void add(int x, int flag) {for (; x <= n; x += (x & (-x))) bits[x] += flag;}
int sum(int x) {int ret = 0; for (; x; x -= (x & (-x))) ret += bits[x]; return ret;}

void solve(int p, int q, int l, int r) {
if (p > q) return;
if (l == r) {
for (int i = p; i <= q; ++i)
if (B[i].op == 3)
ans[B[i].id] = l;
return;
}

int mid = (l + r) >> 1;
for (int i = p; i <= q; ++i) {
if (B[i].op != 3) {
if (B[i].y <= mid)
}
else que[i] = sum(B[i].y) - sum(B[i].x - 1);
}

for (int i = p; i <= q; ++i)
if (B[i].op != 3 && B[i].y <= mid) add(B[i].x, -B[i].op);

q1len = q2len = 0;
for (int i = p; i <= q; ++i)
if (B[i].op != 3) {
if (B[i].y > mid) q2[++q2len] = B[i];
else q1[++q1len] = B[i];
} else {
if (que[i] >= B[i].k) q1[++q1len] = B[i];
else B[i].k -= que[i], q2[++q2len] = B[i];
}

int tmp = p - 1;
for (int i = 1; i <= q1len; ++i)
B[++tmp] = q1[i];
for (int i = 1; i <= q2len; ++i)
B[++tmp] = q2[i];

tmp = p + q1len - 1;
solve(p, tmp, l, mid);
solve(tmp + 1, q, mid + 1, r);
}

int main() {
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; ++i) {
scanf("%d", a + i);
B[++tot] = (node) {i, a[i], 0, 1, 0};
H[++cnt] = a[i];
}

int x, y, k;
char c[2];
for (int i = 1; i <= m; ++i) {
scanf("%s", c);
if (c[0] == 'Q') {
scanf("%d%d%d", &x, &y, &k);
B[++tot] = (node) {x, y, k, 3, ++anstot};
} else {
scanf("%d%d", &x, &y);
B[++tot] = (node) {x, a[x], 0, -1, 0};
B[++tot] = (node) {x, y, 0, 1, 0};
H[++cnt] = (a[x] = y);
}
}

stable_sort(H + 1, H + cnt + 1);
cnt = unique(H + 1, H + cnt + 1) - H;
for (int i = 1; i <= tot; ++i)
if (B[i].op != 3)
B[i].y = lower_bound(H + 1, H + cnt, B[i].y) - H;

solve(1, tot, 1, cnt - 1);

for (int i = 1; i <= anstot; ++i) printf("%d\n", H[ans[i]]);
return 0;
}
posted @ 2017-02-01 12:33  abclzr  阅读(...)  评论(...编辑  收藏