POJ3264 Balanced Lineup（線段樹）

POJ3264 http://poj.org/problem?id=3264

線段樹模版。

題目大意：

Q次區間查詢。

AC代碼：

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
const int maxn = 50000 + 5;
int a[maxn];
int miv[maxn * 4];
int mav[maxn * 4];
int ans1 = -0x3f3f3f;
int ans2 = 0x3f3f3f;
void pushup(int i) {
    miv[i] = min(miv[i << 1], miv[i << 1 | 1]);
    mav[i] = max(mav[i << 1], mav[i << 1 | 1]);
}
void build(int i,int l,int r) {
    if(l == r) {
        miv[i] = a[l];
        mav[i] = a[l];
        return;
    }
    int mid = (l + r) >> 1;
    build(i << 1, l, mid);
    build(i << 1 | 1, mid + 1, r);
    pushup(i);
}
void query(int i, int l, int r, int x, int y) {
    if(x <= l && r <= y) {
        ans1 = max(ans1, mav[i]);
        ans2 = min(ans2, miv[i]);
    }else {
        int mid = (l + r) >> 1;
        if(x <= mid) {
            query(i << 1, l, mid, x, y);
        }
        if(y > mid) {
            query(i << 1 | 1, mid + 1, r, x, y);
        }
    }
}
int main() {
    int N, Q;
    int w, b;
    while(~scanf("%d%d",&N,&Q)) {
        for(int i = 1; i <= N; i++) {
            scanf("%d",&a[i]);
        }
        build(1,1,N);
        while(Q--) {
            ans1 = -0x3f3f3f;
            ans2 = 0x3f3f3f;
            scanf("%d%d",&w,&b);
            query(1,1,N,w,b);
            printf("%d\n",ans1 - ans2);
        }
    }
    return 0;
}