正睿20秋季普转提day1

估分：100+100+30+30=260

实际：100+0+10+10=120

 

T1：

　　正解：简单的模拟

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

typedef long long LL;

LL read()
{
    LL x = 0, f = 0;
    char ch = getchar();
    while (!isdigit(ch)) f = ch == '-', ch = getchar();
    while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    return f ? -x : x;
}

void get(LL& x, LL& y, LL dir)
{
    switch(dir)
    {
    case 0:
        break;
    case 1:
        x = -x;
        swap(x, y);
        break;
    case 2:
        x = -1, y = -1;
        break;
    case 3:
        y = -y;
        swap(x, y);
        break;
    }
}

int main()
{
    LL n = read(), T = read();
    LL x = 0, y = 0, dir = 0;
    for (int i = 1; i <= n; i++)
    {
        LL a = read();
        switch (dir)
        {
        case 1:
            x += a;
            break;
        case 2:
            y += a;
            break;
        case 3:
            x -= a;
            break;
        case 0:
            y -= a;
            break;
        }
        dir += a % 4;
        dir %= 4;
    }


    LL dx = x, dy = y;
    x = y = 0;
    while (T--)
    {
        get(dx, dy, dir);
        x += dx, y += dy;
    }
    printf("%lld", abs(x) + abs(y));
}

 

T2:

　　错题原因：代码写炸了，思路没问题

　　正解：枚举在哪停下，再把之前的餐馆能选则选，优先队列优化一下

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

typedef long long LL;

const int N = 100010;

int n, ans;
LL m, sum;
int t[N];

priority_queue<int> q;

struct Node 
{
    LL x;
    int t;
    Node() {}
} a[N];

inline bool cmp(Node x, Node y) 
{
    return x.x < y.x;
}

int main()
{
    scanf("%d%lld", &n, &m);

    for (int i = 1; i <= n; i++) scanf("%lld%d", &a[i].x, &a[i].t); 

    sort(a + 1, a + 1 + n, cmp);

    LL last = 0;
    for (int i = 1; i <= n; i++)
    {
        LL x = a[i].x;
        int t = a[i].t;
        m -= (x - last);
        last = x;
        while (!q.empty() && sum > m)
        {
            sum -= q.top();
            q.pop();
        }
        if (sum + t <= m) 
        {
            q.push(t);
            sum += t;
        }
        ans = max(ans, (int)q.size());
    }

    printf("%d\n", ans);
}

 

T3：

　　错题原因：想到了这个DP，但不会证明，没敢写，写了暴力，但是写的暴力思路出了点问题

　　正解：L当容量，D当价值跑背包

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

const int N = 1010;

int n;
int f[N], l[N], d[N];

int main() 
{
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", &l[i]);
    for (int i = 1; i <= n; i++) scanf("%d", &d[i]);

    for (int i = 1; i <= n; i++)
    {
        for (int j = n; j >= l[i]; j--)
        {
            f[j] = max(f[j], f[j - l[i]] + d[i]);
        }
    }
    printf("%d", f[n]);
}

 

T4：

　　错题原因：思路想到了，但是set还不熟不会用，不知道怎么优化成满分代码，写了个30分暴力，但是爆int了，只得十分

　　正解：对于每种书用一个权值线段树维护（或set），对于总的再用一个权值线段树维护（或set），没有选的书也要维护一下，RETURN操作时要用，再模拟一下即可

 

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

const int N = 300010;

inline int read()
{
    int s = 0; char ch = getchar();
    while (ch < '0' || ch>'9') ch = getchar();
    while (ch <= '9' && ch >= '0') { s = (s << 1) + (s << 3) + (ch ^ 48); ch = getchar(); }
    return s;
}

int n, m;
long long ans;
multiset<int> s1[N], s2[N], f, del;

inline void add(int x)
{
    if (f.size() < n)
    {
        f.insert(x);
        ans += x;
    }
    else
    {
        int flag = *f.begin();
        if (x > flag)
        {
            del.insert(flag);
            ans -= flag;
            f.erase(f.begin());
            f.insert(x);
            ans += x;
        }
        else del.insert(x);
    }
}

inline void modify(int x)
{
    if (f.find(x) != f.end())
    {
        ans -= x;
        f.erase(f.find(x));
        if (!del.empty())
        {
            int flag = *del.rbegin();
            del.erase(--del.end());
            f.insert(flag);
            ans += flag;
        }
    }
    else del.erase(x);
}

int main()
{
    scanf("%d%d", &n, &m);

    for (int i = 1; i <= m; i++)
    {
        char opt[10];
        scanf(" %s", opt);
        int t = read(), x = read();
        if (opt[0] == 'B')
        {
            if (s1[t].size() < t)
            {
                s1[t].insert(x);
                add(x);
            }
            else
            {
                int flag = *s1[t].begin();
                if (x > flag)
                {
                    modify(flag);
                    s2[t].insert(flag);
                    s1[t].erase(s1[t].begin());
                    add(x);
                    s1[t].insert(x);
                }
                else s2[t].insert(x);
            }
        }
        else
        {
            if (s1[t].find(x) != s1[t].end())
            {
                s1[t].erase(s1[t].find(x));
                modify(x);
                if (!s2[t].empty())
                {
                    int flag = *s2[t].rbegin();
                    s1[t].insert(flag);
                    s2[t].erase(--s2[t].end());
                    add(flag);
                }
            }
            else s2[t].erase(s2[t].find(x));
        }
        printf("%lld\n", ans);
    }
}

 

总结：

　　题目的正解思路都是想到了的，但不是写炸了就是不会写，代码能力还是太差，以前学的一些数据结构忘了，要复习一下，数据结构还要多写，容易写错；题目的数据范围也要注意，注意会不会爆int什么的