Codeforces Round #424 (Div. 2) D 思维 E set应用,树状数组

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Codeforces Round #424 (Div. 2, rated, based on VK Cup Finals)

D. Office Keys

题意:一条直线上,有个办公室坐标 p,有 n个人在a[i],有 k把钥匙在b[i],每个人必须拿到一把钥匙,然后到办公室。问怎么安排花的时间最短。

tags:还是不懂套路啊。。其实多画两下图就能够感觉出来,2333

关键是要看出来,n个人拿的 n把钥匙应该是连续的。

然后,就是瞎暴力。。

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define rep(i,a,b) for (int i=a;i<=b;i++)
#define per(i,b,a) for (int i=b;i>=a;i--)
#define mes(a,b)  memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define MP make_pair
#define PB push_back
#define fi  first
#define se  second
typedef long long ll;
const int N = 2005;

int n, k;
ll  p, a[N], b[N];
int main()
{
    scanf("%d %d %lld", &n, &k, &p);
    rep(i,1,n) scanf("%lld", &a[i]);
    rep(i,1,k) scanf("%lld", &b[i]);
    sort(a+1, a+1+n);
    sort(b+1, b+1+k);
    ll  ans=1e18;
    for(int cb=1; cb+n-1<=k; ++cb)
    {
        ll ans1=0;
        rep(i,1,n)
        {
            ans1 = max(ans1, abs(a[i]-b[cb+i-1])+abs(b[cb+i-1]-p));
        }
        ans=min(ans, ans1);
    }
    printf("%lld\n", ans);

    return 0;
}

 E. Cards Sorting

题意:n 个数,操作:每次取出头一个数,如果是这些数中最小的就删除,如果不是就放到底部。 问要多少次操作才能全部删除这 n个数。

tags:利用set和树状数组,快速模拟操作。一开始思路错了,想的是for一遍,每个数的会对前面比它大的数产生贡献,最后发现不行,白折腾半小时,心累

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define rep(i,a,b) for (int i=a;i<=b;i++)
#define per(i,b,a) for (int i=b;i>=a;i--)
#define mes(a,b)  memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define MP make_pair
#define PB push_back
#define fi  first
#define se  second
typedef long long ll;
const int N = 100005;

struct Bit
{
    int bi[N];
    void add(int x, int y) { for(; x<N; x+=x&-x) bi[x]+=y; }
    ll  sum(int x) { ll ans=0; for(; x>0; x-=x&-x) ans+=bi[x]; return ans; }
}bit;
int n, a[N];
set<int > se[N];
int main()
{
    scanf("%d", &n);
    rep(i,1,n)
    {
        scanf("%d", &a[i]);
        se[a[i]].insert(i);
        bit.add(i, 1);
    }
    ll  ans=0;
    int now=1, mi=1;
    rep(ci,1,n)
    {
        while(se[mi].empty()) ++mi;
        if(se[mi].lower_bound(now)!=se[mi].end())
        {
            int pos=now;
            now = *se[mi].lower_bound(now);
            ans += bit.sum(now)-bit.sum(pos-1);
        }
        else
        {
            int pos=now;
            now = *se[mi].lower_bound(1);
            ans += bit.sum(n)-bit.sum(pos-1)+bit.sum(now);
        }
        se[mi].erase(now);
        bit.add(now, -1);
    }
    printf("%lld\n", ans);

    return 0;
}

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