POJ 2796 Feel Good 单调栈或者笛卡尔树

题目：

http://poj.org/problem?id=2796

题意：

给定一个长度为n<script type="math/tex" id="MathJax-Element-27">n</script>的数组，一个区间的值为这个区间的中所有值的和与最小值的乘积，求区间的最大值，并输出区间的左右端点

思路：

明显的单调栈，笛卡尔树也可以做

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long ll;
const int N = 100000 + 10, INF = 0x3f3f3f3f;

int a[N];
ll sum[N];

int top, stk[N];
int l[N], r[N];

int main()
{
    int n;
    while(~ scanf("%d", &n))
    {
        for(int i = 1; i <= n; i++) scanf("%d", &a[i]), sum[i] = sum[i-1] + a[i];
        top = 0;
        for(int i = 1; i <= n; i++)
        {
            while(top > 0 && a[stk[top-1]] >= a[i]) top--;
            l[i] = (top == 0) ? 1 : stk[top-1] + 1;
            stk[top++] = i;
        }
        top = 0;
        for(int i = n; i >= 1; i--)
        {
            while(top > 0 && a[stk[top-1]] >= a[i]) top--;
            r[i] = (top == 0) ? n : stk[top-1] - 1;
            stk[top++] = i;
        }
        ll ans = -1;
        int tl, tr;
        for(int i = 1; i <= n; i++)
        {
            ll tmp = (sum[r[i]] - sum[l[i]-1]) * a[i];
            if(tmp > ans)
            {
                ans = tmp, tl = l[i], tr = r[i];
            }
        }
        printf("%lld\n", ans);
        printf("%d %d\n", tl, tr);
    }
    return 0;
}

笛卡尔树：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef pair<int, int> pii;
typedef long long ll;
const int N = 100000 + 10, INF = 0x3f3f3f3f;

struct node
{
    int val, pri, fat, id, son[2];
    friend bool operator< (node a, node b)
    {
        return a.val < b.val;
    }
    void init(int _val, int _pri, int _fat, int _id)
    {
        val = _val, pri = _pri, fat = _fat, id = _id;
        son[0] = son[1] = 0;
    }
}tr[N];
int top, stk[N];
int root;
int a[N];
ll sum[N];
ll ans;
int ansl, ansr;
int cartesian_build(int n)
{
    top = 0;
    for(int i = 1; i <= n; i++)
    {
        int k = top;
        while(k > 0 && tr[stk[k-1]].pri >= tr[i].pri) k--;
        if(k != 0)
        {
            tr[stk[k-1]].son[1] = i;
            tr[i].fat = stk[k-1];
        }
        if(k != top)
        {
            tr[i].son[0] = stk[k];
            tr[stk[k]].fat = i;
        }
        stk[k++] = i;
        top = k;
    }
    return stk[0];
}
int dfs(int x, int f)
{
    if(! x) return 0;
    int l = dfs(tr[x].son[0], 0);
    int r = dfs(tr[x].son[1], 1);
    if(l == 0) l = tr[x].id; //说明此节点管辖区间的左端点是自身
    if(r == 0) r = tr[x].id; //同理
    ll tmp = (sum[r] - sum[l-1]) * tr[x].pri;
    if(tmp > ans) ans = tmp, ansl = l, ansr = r;
    return f ? r : l; //判断返回右端点还是左端点
}
int main()
{
    int n;
    while(~ scanf("%d", &n))
    {
        tr[0].init(0, 0, 0, 0);
        for(int i = 1; i <= n; i++)
        {
            scanf("%d", &a[i]);
            sum[i] = sum[i-1] + a[i];
            tr[i].init(i, a[i], 0, i);
        }
        sort(tr + 1, tr + 1 + n);
        root = cartesian_build(n);
        ans = -1;
        dfs(root, 1);
        printf("%lld\n", ans);
        printf("%d %d\n", ansl, ansr);
    }
    return 0;
}