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POJ - 3264 Balanced Lineup(线段树)


题目大意:给你N个数,然后询问,询问的是区间内最大值和最小值的差

解题思路:线段树的裸题了,维护最大值和最小值即可

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int INF = 0x3f3f3f3f;
const int N = 50010 << 2;

int n, q;
int val[N], Min[N], Max[N];

void PushUp(int u) {
    Max[u] = max(Max[u << 1], Max[u << 1 | 1]);
    Min[u] = min(Min[u << 1], Min[u << 1 | 1]);
}

void build(int u, int l, int r) {
    if (l == r) {
        Max[u] = Min[u] = val[l];
        return ;
    }
    int mid = (l + r) >> 1;
    build(u << 1, l, mid);
    build(u << 1 | 1, mid + 1, r);
    PushUp(u);
}

int query(int u, int l, int r, int L, int R, int flag) {
    if (L <= l && r <= R) {
        if (flag) return Max[u];
        else return Min[u];
    }
    int mid = (l + r) >> 1;
    int ans = INF;
    if (flag) ans = -INF;

    if (L <= mid) {
        if (flag) ans = max(ans, query(u << 1, l, mid, L, R, flag));
        else ans = min(ans, query(u << 1, l, mid, L, R, flag));
    }

    if (R > mid) {
        if (flag) ans = max(ans, query(u << 1 | 1, mid + 1, r, L, R, flag));
        else ans = min(ans, query(u << 1 | 1, mid + 1, r, L, R, flag));
    }
    return ans;
}

void solve() {
    for (int i = 1; i <= n; i++)
        scanf("%d", &val[i]);
    build(1, 1, n);
    int a, b;
    while (q--) {
        scanf("%d%d", &a, &b);
        printf("%d\n", query(1, 1, n, a, b, 1) - query(1, 1, n, a, b, 0));
    }
}

int main() {
    while (scanf("%d%d", &n, &q) != EOF) solve();
    return 0;
}


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