HDU 5649 線段樹+二分

題意：

題目鏈接：http://acm.split.hdu.edu.cn/showproblem.php?pid=5649
n個數的排列，兩種操作，一種將[L,R]區間內的數遞減排序，另一種將[L,R]中的數遞增排序，問最後第k個位置的數字是多少。

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思路：

好題。
因爲最後考慮的只是第k個位置的數字，二分答案。
每次將序列中比x大的都標記爲1，其餘包括x標記爲0，每次排序前，先查詢區間中有多少個1和0，然後按照操作將1和0分別更新到區間的左右兩部分，用線段樹區間更新區間查詢即可。判斷時如果第k個位置爲0則r=m-1,否則l=m+1。

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代碼：

#include <bits/stdc++.h>
using namespace std;
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
const int MAXN = 1e5 + 10;

int n, q, k;
int a[MAXN], b[MAXN], op[MAXN], LL[MAXN], RR[MAXN];
int C[MAXN << 2], lazy[MAXN << 2];

void pushUp(int rt) {
    C[rt] = C[rt << 1] + C[rt << 1 | 1];
}

void pushDown(int rt, int len) {
    if (lazy[rt] != -1) {
        lazy[rt << 1] = lazy[rt << 1 | 1] = lazy[rt];
        C[rt << 1] = (len - (len >> 1)) * lazy[rt];
        C[rt << 1 | 1] = (len >> 1) * lazy[rt];
        lazy[rt] = -1;
    }
}

void build(int l, int r, int rt) {
    lazy[rt] = -1;
    if (l == r) {
        C[rt] = a[l];
        return;
    }
    int m = (l + r) >> 1;
    build(lson);
    build(rson);
    pushUp(rt);
}

void update(int L, int R, int v, int l, int r, int rt) {
    if (L <= l && r <= R) {
        C[rt] = v * (r - l + 1);
        lazy[rt] = v;
        return;
    }
    pushDown(rt, r - l + 1);
    int m = (l + r) >> 1;
    if (L <= m) update(L, R, v, lson);
    if (R > m) update(L, R, v, rson);
    pushUp(rt);
}

int query(int L, int R, int l, int r, int rt) {
    if (L <= l && r <= R) return C[rt];
    pushDown(rt, r - l + 1);
    int m = (l + r) >> 1, res = 0;
    if (L <= m) res += query(L, R, lson);
    if (R > m) res += query(L, R, rson);
    return res;
}

int cal(int x) {
    for (int i = 1; i <= n; i++) {
        a[i] = (b[i] > x);
    }
    build(1, n, 1);
    //cout << C[1] << endl;
    for (int i = 1; i <= q; i++) {
        int L = LL[i], R = RR[i];
        int one = query(L, R, 1, n, 1);
        //cout << "(" << L << ", " << R << ") one = " << one << endl;
        int zero = R - L + 1 - one;
        if (op[i] == 0) {
            if (zero > 0) update(L, L + zero - 1, 0, 1, n, 1);
            if (one > 0) update(R - one + 1, R, 1, 1, n, 1);
        }
        else {
            if (one > 0) update(L, L + one - 1, 1, 1, n, 1);
            if (zero > 0) update(R - zero + 1, R, 0, 1, n, 1);
        }
    }
    return  (query(k, k, 1, n, 1) == 0);
}

void solve(int l, int r) {
    int res = -1;
    while (l <= r) {
        int mid = (l + r) >> 1;
        int tmp = cal(mid);
        //cout << mid << " " << tmp << endl;
        if (tmp) {
            r = mid - 1;
            res = mid;
        }
        else l = mid + 1;
    }
    printf("%d\n", res);
}

int main() {
    //freopen("in.txt", "r", stdin);
    int T;
    scanf("%d", &T);
    while (T--) {
        scanf("%d%d", &n, &q);
        for (int i = 1; i <= n; i++) {
            scanf("%d", &b[i]);
        }
        for (int i = 1; i <= q; i++)
            scanf("%d%d%d", &op[i], &LL[i], &RR[i]);
        scanf("%d", &k);
        solve(1, n);
    }
    return 0;
}