1128

题目链接：http://ifrog.cc/acm/problem/1128

思路有两种：

• 线段树查区间最小值和最大值
• RMQ查区间最小值和最大值

线段树代码：

#include<bits/stdc++.h>
using namespace std;

const long long INF = 0x3f3f3f3f;
long long n;
long long a[100007], b[100007];
struct node
{
    long long l, r, x, y;
};
node tree[100007 << 2];

void build(long long l, long long r, long long k)
{
    tree[k].l = l, tree[k].r = r;
    if(l == r)
    {
        tree[k].x = tree[k].y = a[l];
        return;
    }
    long long mid = (l + r) >> 1;
    build(l, mid, k << 1);
    build(mid+1, r, k << 1|1);
    tree[k].x = max(tree[k<<1].x, tree[k<<1|1].x);
    tree[k].y = min(tree[k<<1].y, tree[k<<1|1].y);
}

long long query_x(long long l, long long r, long long k)
{
    if(tree[k].l >= l && tree[k].r <= r)
    {
        return tree[k].x;
    }

    long long res = -1e10-7;

    if(tree[k<<1].r >= l)
        res = max(res, query_x(l, r, k<<1));
    if(tree[k<<1|1].l <= r)
        res = max(res, query_x(l, r, k<<1|1));
    return res;
}

long long query_y(long long l, long long r, long long k)
{
    if(tree[k].l >= l && tree[k].r <= r)
    {
        return tree[k].y;
    }

    long long res = 1e10+7;

    if(tree[k<<1].r >= l)
        res = min(res, query_y(l, r, k<<1));
    if(tree[k<<1|1].l <= r)
        res = min(res, query_y(l, r, k<<1|1));
    return res;
}

int main()
{
    while(scanf("%d", &n) != EOF)
    {
        for(long long i=1; i<=n; ++ i)
            scanf("%lld", &a[i]);
        for(long long i=1; i<=n; ++ i)
            scanf("%lld", &b[i]);
        build(1, n, 1);
        for(long long i=1; i<=n; ++ i)
        {
            long long x = query_x(i-b[i]+1, i, 1);
            long long y = query_y(i-b[i]+1, i, 1);
            printf("%lld
", x * y);
        }
    }
    return 0;
}

RMQ代码

#include<bits/stdc++.h>
using namespace std;

int a[100007], b[100007];
int mx[100007][20], mn[100007][20];

void RMQ(int num)
{
    memset(mx, 0, sizeof(mx));
    memset(mn, 0, sizeof(mn));
    for(int i=1; i<=num; ++ i)
        mx[i][0] = mn[i][0] = a[i];
    for(int j=1; j<=20; ++ j)
    {
        for(int i=1; i<=num; ++ i)
        {
            if(i + (1 << j) - 1 <= num)
            {
                mx[i][j] = max(mx[i][j-1], mx[i+(1 << (j-1))][j-1]);
                mn[i][j] = min(mn[i][j-1], mn[i+(1 << (j-1))][j-1]);
            }
        }
    }
}

int main()
{
    int n;
    while(scanf("%d", &n) != EOF)
    {
        for(int i=1; i<=n; ++ i)
            scanf("%d", &a[i]);
        for(int i=1; i<=n; ++ i)
            scanf("%d", &b[i]);
        RMQ(n);
        for(int i=1; i<=n; ++ i)
        {
            int t = (log10(b[i]) / log10(2));
            long long x = max(mx[i-b[i]+1][t], mx[i-(1<<t)+1][t]);
            long long y = min(mn[i-b[i]+1][t], mn[i-(1<<t)+1][t]);
            printf("%lld
", x*y);
        }
    }
    return 0;
}