8.5 Round 2

三道思维题，考场上我真是一道也不会啊

 

T1 装饰大楼：https://www.luogu.org/problem/T92137

失误：脑子貌似压根没转，没认真考虑怎么判重，乱写了个哈希还挂了...

考虑位于位置i，前面的最大值是k，那么这个位置可以填1-k+1

判重：如果a[i+1]值在[1,k+1]，那么会有一种重复情况，ans--即可

无解：1.前面是1-k，这里出现k+3及以上

　　    2.有两个位置出现k+2

特殊的，当只有一个位置pos出现k+2时，我们只能在1-pos-1填入一个大小为k+1的数，且填入位置之前最大值必须为k

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<climits>
#include<map>
#include<queue>
using namespace std;
#define O(x) cout << #x << " " << x << endl;
#define B cout << "breakpoint" << endl;
#define clr(a) memset(a,0,sizeof(a));
#define int long long
typedef long long ll;
inline int read()
{
    int ans = 0,op = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9')
    {
        if(ch == '-') op = -1;
        ch = getchar();
    }
    while(ch >= '0' && ch <= '9')
    {
        (ans *= 10) += ch - '0';
        ch = getchar();
    }
    return ans * op;
}
const int maxn = 1e6 + 5;
int n,a[maxn],ans;
main()
{
    freopen("A.in","r",stdin);
    freopen("A.out","w",stdout);
    n = read(); n--;
    for(int i = 1;i <= n;i++) 
        a[i] = read();
    int mx = 0;
    int pos = 0;
    for(int i = 1;i <= n;i++)
    {
        if(a[i] > mx + 2) { printf("0\n"); return 0; }
        else if(a[i] == mx + 2)
        {
            if(pos) { printf("0\n"); return 0; }
            pos = a[i];
        }
        mx = max(mx,a[i]);
    }
    mx = 0;
    for(int i = 0;i <= n;i++)
    {
        mx = max(mx,a[i]);
        if(pos)
        {
            if(mx == pos - 2) ans++;
            continue;
        }
        ans += mx + 1;
        if(a[i + 1] <= mx + 1 && a[i + 1] > 0) ans--;
    }
    printf("%lld",ans);
}

 

T2：https://www.luogu.org/problem/T92119

将所有端点拿出来，排序后离散化

对于相邻两个点，可以根据进出情况分为四种

将一种依赖的情况连边，dfs获得新的序列

f[i][j][0]/[1]表示前i个点，带j把钥匙，第i个人带或不带的最大答案

O（n^2）

依旧是没有思考...

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<climits>
#include<map>
#include<queue>
using namespace std;
#define O(x) cout << #x << " " << x << endl;
#define B cout << "breakpoint" << endl;
#define clr(a) memset(a,0,sizeof(a));
#define int long long
typedef long long ll;
const ll inf = LONG_LONG_MAX - 10;
inline int read()
{
    int ans = 0,op = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9')
    {
        if(ch == '-') op = -1;
        ch = getchar();
    }
    while(ch >= '0' && ch <= '9')
    {
        (ans *= 10) += ch - '0';
        ch = getchar();
    }
    return ans * op;
}
const int maxn = 2005;
struct node
{
    int x,id;
    bool in,out;
    bool operator < (const node &b) const
    {
        return x < b.x;
    }
}a[maxn << 1];
int n,tot,m,k,ans;
int to[maxn],p[maxn],g[maxn],f[maxn][maxn][2],in[maxn];
int t[maxn],cnt;
bool vis[maxn];
void dfs(int u)
{
    if(!u) return;
    in[u] = 1;
    t[++cnt] = u;
    dfs(to[u]);
}
main()
{
    freopen("B.in","r",stdin);
    freopen("B.out","w",stdout);
    n = read(),m = read(),k = read();
    for(int i = 1;i <= n;i++)
    {
        a[++tot].x = read(),a[tot].id = i,a[tot].out = 1,a[tot].in = 0;
        a[++tot].x = read(),a[tot].id = i,a[tot].in = 1,a[tot].out = 0;
    }
    sort(a + 1,a + 1 + n * 2);
    ans += a[1].x + m - a[n * 2].x;
    for(int i = 1;i <= n * 2 - 1;i++)
    {
        int x1 = a[i].id,x2 = a[i + 1].id;
        int tp = a[i + 1].x - a[i].x;
        if(a[i].in && a[i + 1].out) ans += tp;//3
        if(a[i].in && a[i + 1].in) p[x2] += tp;//4
        if(a[i].out && a[i + 1].out) p[x1] += tp;//1
        if(a[i].out && a[i + 1].in) { if(x1 == x2) p[x1] += tp;
                                    else in[x2] = 1,g[x2] = tp,to[x1] = x2; }
    }
    for(int i = 1;i <= n;i++) 
        if(!in[i]) dfs(i);
    for(int i = 0;i <= n;i++)    
        for(int j = 0;j <= k;j++)
            f[i][j][0] = f[i][j][1] = -inf;
    f[0][0][0] = 0;
    for(int i = 1;i <= n;i++)
    {
        for(int j = 0;j <= k;j++)
        {
            f[i][j][0] = max(f[i - 1][j][0],f[i - 1][j][1]);
            if(j) f[i][j][1] = max(f[i - 1][j - 1][1] + g[t[i]],f[i - 1][j - 1][0]) + p[t[i]];
        }
    }
    printf("%lld",ans + max(f[n][k][1],f[n][k][0]));
}

            

 

T3：咕