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POJ - 3264 Balanced Lineup(线段树)

2015-10-20 23:33 369 查看
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
, Min
, Max
;

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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