【51nod 1287 加农炮】线段树

最新推荐文章于 2021-10-17 11:16:18 发布

原创最新推荐文章于 2021-10-17 11:16:18 发布 · 114 阅读

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博客介绍了51nod1287问题的解法，以元素下标为节点存储高度值，每次找出大于等于当前高度的第一个高度的下标值，将该下标前一个数的高度加一，使用线段树维护，时间复杂度为O(nlogn+Qlogn)。

51nod1287
其实就是以他们的下标为节点存储他们的高度值
然后每次找大于等于他的第一个高度的下标值把这个下标前面那个数高度加一
然后用线段树维护就行了
时间复杂度是是 O(nlogn+Qlogn)

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
#define mid ((l+r)>>1)

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);

/*namespace sgt
{
    #define mid ((l+r)>>1)

    #undef mid
}*/
#define N 1000000
int Max(int a,int b){if(a>b) return a;return b;}
const int MAX_N = 50025;
int arr[MAX_N],q[MAX_N],maxx[MAX_N<<2];
void up(int rt)
{
    maxx[rt] = Max(maxx[rt<<1],maxx[rt<<1|1]);
}
void build(int rt,int l,int r)
{
    maxx[rt] = -1;
    if(l==r)
    {
        maxx[rt] = arr[l];
        return;
    }
    build(rt<<1,l,mid);
    build(rt<<1|1,mid+1,r);
    up(rt);
}
void update(int rt,int l,int r,int x)
{
    if(l==r)
    {
        maxx[rt]++;
        return ;
    }
    if(x<=mid) update(rt<<1,l,mid,x);
    else update(rt<<1|1,mid+1,r,x);
    up(rt);
}
int query(int rt,int l,int r,int x,int y,int v)
{
    if(maxx[rt]<v) return -1;
    if(l==r)
    {
        if(maxx[rt]>=v) return l;
        return -1;
    }
    int ans = -1;
    if(maxx[rt<<1]>=v&&x<=mid) ans = query(rt<<1,l,mid,x,y,v);
    if(ans==-1&&mid<y) ans = query(rt<<1|1,mid+1,r,x,y,v);
    return ans;
}
int query_(int rt,int l,int r,int x,int y)
{
    if(x<=l&&r<=y)
    {
        return maxx[rt];
    }
    if(x>mid) return query_(rt<<1|1,mid+1,r,x,y);
    else if(y<=mid) return query_(rt<<1,l,mid,x,y);
    else return Max(query_(rt<<1,l,mid,x,y),query_(rt<<1|1,mid+1,r,x,y));
}
void debug(int rt,int l,int r)
{
    if(l==r)
    {
        printf("%d\n",maxx[rt]);
        return ;
    }
    debug(rt<<1,l,mid);
    debug(rt<<1|1,mid+1,r);
}
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int n,Q,d;
    scanf("%d%d",&n,&Q);
    for(int i = 1;i<=n;++i) scanf("%d",&arr[i]);
    build(1,1,n);
    while(Q--)
    {
        scanf("%d",&d);
        if(d>maxx[1]||d<=query_(1,1,n,1,1)) continue;
        int xb = query(1,1,n,1,n,d);
        update(1,1,n,xb-1);
    }
    debug(1,1,n);
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}