植树节，种种小树苗就好了。

Segment Tree(TYVJ P1038)

没有更新的区间最值查询，够小（简单）了吧

#include<cstdio>
#include<algorithm>
using namespace std;
#define MAXM 100000
#define lson l , m , rt<<1
#define rson m+1 , r , rt<<1|1

int minv[MAXM<<2],n,m;


void pushup(int rt)
{
    minv[rt] = min(minv[rt<<1] , minv[rt<<1|1]);
}


void build(int l,int r,int rt)
{
    if(l == r){
        scanf("%d",&minv[rt]);
        return ;
    }
    int m = (l+r) >> 1;
    build(lson);
    build(rson);
    pushup(rt);
}


int query(int L,int R,int l,int r,int rt)
{
    if(L<=l && r<=R) return minv[rt];
    int m = (l+r) >> 1;
    int ret = 2e9;
    if(L <= m) ret = min(ret , query(L,R,lson));
    if(R > m) ret = min(ret , query(L,R,rson));
    return ret;
}


void work()
{
    int a,b;
    for(int i = 1 ; i <= n ; i++)
    {
        scanf("%d%d",&a,&b);
        printf("%d%s",query(a,b,1,m,1) , i<n ? " ":"
");
    }
}


void init()
{
    scanf("%d%d",&m,&n);
    build(1,m,1);
}


int main()
{
    init();
    work();
    return 0;
}

Fenwick Tree(TYVJ P1432)

求数组中i<j<k，a[i]>a[j]<a[k] 和 a[i]<a[j]>a[k] 分别的个数

用Fenwick来求数组前i个数中比a[i]小的个数就好了，够小（简单）了吧

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

#define MAXN 200010
#define LL long long

int n,a[MAXN];
int b[MAXN],x[MAXN];


int sum(int k)
{
    int ret = 0;
    while(k > 0)
    {
        ret += b[k];
        k -= k&-k;
    }
    return ret;
}


void add(int k)
{
    while(k <= n)
    {
        b[k]++;
        k += k&-k;
    }
}


void work()
{
    LL ans1 = 0,ans2 = 0;
    for(int i = 1 ; i <= n ; i++)
    {
        x[i] = sum(a[i]-1);
        add(a[i]);
    }
    for(int i = 2 ; i < n ; i++)
    {
        ans1 += (i - 1 - x[i]) * (n - i - a[i] + 1 + x[i]);
        ans2 += x[i] * (a[i] - 1 - x[i]);
    }
    printf("%lld %lld
",ans1 , ans2);
}


void init()
{
    scanf("%d",&n);
    for(int i = 1 ; i <= n ; i++) scanf("%d",&a[i]);
}


int main()
{
    init();
    work();
    return 0;
}

Balanced Search Tree(TYVJ P1185)

求数组中Σ(i=1..n)min{abs(a[i]-a[j])|j<i}

用STL中自带的红黑树容器set求比a[i]小的最大数和比a[i]大的最小数就好了，够小（简单）了吧

#include<cstdio>
#include<set>
#include<algorithm>
using namespace std;

int n;


set<int>::iterator it;
int work()
{
    int ans,maxv,minv,v,a,b;
    scanf("%d",&v);
    maxv = minv = ans = v;
    set<int> S,S1;
    S.insert(v);
    S1.insert(-v);
    for(int i = 1 ; i < n ; i++)
    {
        a = b = -1;
        scanf("%d",&v);
        if(maxv >= v)
        {
            it = S.lower_bound(v);
            a = *it - v;
        }
        if(minv <= v)
        {
            it = S1.lower_bound(-v);
            b = v + *it;
        }
        S.insert(v);
        S1.insert(-v);
        maxv = max(maxv , v);
        minv = min(minv , v);
        if(a>=0 && b>= 0) ans += min(a , b);
        else ans += (a>=0 ? a : b);
    }
    printf("%d
",ans);
}


int main()
{
    scanf("%d",&n);
    work();
    return 0;
}

Suffix Tree()

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