2024.11.13

今日总结
1：http://www.nfls.com.cn:20035/contest/2085/problem/1
这道题考察了字符串哈希，没学过，今天上午学习了字符串哈希

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#include <bits/stdc++.h>
using namespace std;
const long long p = 1e9 + 7;
string s;
unsigned long long int base[1000005], h1[1000005], h2[1000005], sum = 0;
int n;
bool check(int x, int d)
{
    int l = x - d + 1, r = x + d;
    if (h1[r] - h1[l - 1] * base[r - l + 1] == h2[l] - h2[r + 1] * base[r - l + 1])
    {
        return true;
    }
    return false;
}
int main()
{
    cin >> n;
    cin >> s;
    s = ' ' + s;
    for (int i = 1; i <= n; i++)
    {
        h1[i] = h1[i - 1] * p + (s[i] == '1');
    }
    for (int i = n; i >= 1; i--)
    {
        h2[i] = h2[i + 1] * p + (s[i] == '0');
    }
    base[0] = 1;
    for (int i = 1; i <= n; i++)
    {
        base[i] = base[i - 1] * p;
    }
    for (int i = 1; i < n; i++)
    {
        int l = 0, r = min(i, n - i), mid, ans;
        while (l <= r)
        {
            mid = (l + r) >> 1;
            if (check(i, mid))
            {
                ans = mid, l = mid + 1;
            }
            else
            {
                r = mid - 1;
            }
        }
        sum += ans;
    }
    cout << sum;
    return 0;
}

2：http://www.nfls.com.cn:20035/contest/2090/problem/6
这道题是一道树形Dp的题目，主要复习了树形Dp的Dp建立含义，转移的方式

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//dp[i][j]是以i为根的子树操作j次后满足所需能剩下的最多金币。
#include<bits/stdc++.h>

using namespace std;

const int N = 2010;
const int INF = 0x3f3f3f3f;

int n,x;
int val[N],siz[N],dp[N][N],f[N];
vector<int> vec[N];

void dfs(int u,int fa)
{
    dp[u][0]  = x - val[u];//剩余金币数量
    siz[u] = 1;
    for(int i = 0;i < vec[u].size();i ++)
    {
        int v = vec[u][i];
        if(v == fa) continue;
        dfs(v,u);
        siz[u] += siz[v];
        memset(f,-0x3f,sizeof(f));
        for(int j = siz[u] - 1;j >= 0;j --)
        {
            for(int k = siz[v] - 1;k >= 0;k --)
            {
                if(j - k - 1 >= 0) f[j] = max(f[j],dp[v][k] + dp[u][j - k - 1]);
                if(dp[v][k] >= 0 && j - k >= 0) f[j] = max(f[j],dp[u][j - k]);
            }
        }
        for(int j = siz[u];j >= 0;j --) dp[u][j] = f[j];
    }
}

int main()
{
    scanf("%d%d",&n,&x);
    for(int i = 1;i <= n;i ++)  
        scanf("%d",&val[i]);
    for(int i = 1;i < n;i ++)
    {
        int u,v;
        scanf("%d%d",&u,&v);
        vec[u].push_back(v);
        vec[v].push_back(u);
    }
    memset(dp,-0x3f,sizeof(dp));
    dfs(1,0);
    int ans = INF;
    for(int i = 0;i < n;i ++)
    {
        if(dp[1][i] >= 0)
        {
            printf("%d\n",i);
            return 0;
        }
    }
    return 0;
}

3：http://www.nfls.com.cn:20035/contest/2082/problem/1
这道题是一道简单的贪心，与排队接水的贪心思路一样

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#include<bits/stdc++.h>

using namespace std;

const int N = 5e4 + 10;

int n,m,d,l,id,ans;
int sp[N],total[N];

int main()
{
    scanf("%d%d%d%d",&n,&m,&d,&l);
    for(int i = 1;i <= n;i ++)
        scanf("%d",&sp[i]);
    sort(sp + 1,sp + 1 + n);
    int x = 1;
    while(x <= n)
    {
        while(sp[x] - total[id] * d < l && x <= n) x ++;
        if(x > n) break;
        ans ++;
        total[id] ++;
        id ++;
        x ++;
        if(id == m) id = 0;
    }
    printf("%d\n",ans);
    return 0;
}

4：http://www.nfls.com.cn:20035/contest/2090/problem/5#
这道题是一个状压Dp,复习了状压Dp状态的表示和转移方法，将其想象为可以和不可以

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#include<bits/stdc++.h>

using namespace std;

const int N = 450;

int n,m,cnt;
int head[16],nxt[N],to[N],calc[1 << 16],dp[16][1 << 16][8];

void dfs(int id,int s,int h)
{
    if(dp[id][s][h] != -1) return;
    if((1 << h) < calc[s])
    {
        dp[id][s][h] = 0;
        return;
    }
    if(s == (1 << id))
    {
        dp[id][s][h] = 1;
        return;
    }
    if(h == 0)
    {
        dp[id][s][h] = 0;
        return;
    }
    dp[id][s][h] = 0;
    int i,j;
    for(i = s & (s - 1);i;i = (i - 1) & s)
    {
        if((1 << id) & i) continue;
        for(j = head[id];j != -1;j = nxt[j])
        {
            if(!((1 << to[j]) & s)) continue;
            if(!((1 << to[j]) & i)) continue;
            dfs(id,s - i,h - 1),dfs(to[j],i,h - 1);
            dp[id][s][h] += dp[id][s - i][h - 1] * dp[to[j]][i][h - 1];
        }
    }
}

int main()
{
    int i,j,id,num;
    while(scanf("%d %d",&n,&m) != EOF)
    {
        memset(head,-1,sizeof(head));
        memset(dp,-1,sizeof(dp));
        memset(calc,0,sizeof(calc));
        cnt = 0,id = 0,num = 1;
        while(num < n)
        {
            num *= 2;
            id ++;
        }
        m --;
        for(i = 0;i < (1 << n);i ++)
            for(j = 0;j < n;j ++)
                if((1 << j) & i) calc[i] ++;
        for(i = 0;i < n;i ++)
        {
            for(j = 0;j < n;j ++)
            {
                scanf("%d",&num);
                if(num == 1)
                {
                    nxt[cnt] = head[i];
                    to[cnt] = j;
                    head[i] = cnt;
                    cnt ++;
                }
            }
        }
        dfs(m,(1 << n) - 1,id);
        printf("%d\n",dp[m][(1 << n) - 1][id]);
    }
    return 0;
}