正睿20NOIp前冲刺day11

估分：100+100+10+10=220

实际：100+20+10+10=140

T1:

　　原题

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

inline int read()
{
    int 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;
}

const int N = 1000010;
const int INF = 0x3f3f3f3f;

int n;
int dep[N], f[N], rev[N];
int h[N], num[N << 1], nex[N << 1], dqx;

void add(int a, int b)
{
    num[dqx] = b;
    nex[dqx] = h[a];
    h[a] = dqx++;
}

void dfs(int u, int pa)
{
    int cnt = 0, res = 0;
    for (int i = h[u]; ~i; i = nex[i])
    {
        int j = num[i];
        if (j == pa) continue;
        dep[j] = dep[u] + 1;
        dfs(j, u);
        f[u] = f[u] + f[j];
        if (rev[j] + 1 <= dep[u])
        {
            f[u] += rev[j] + 1 - dep[u];
            rev[u] = max(rev[u], rev[j] + 1);
            cnt++;
        }
        res++;
    }
    if (cnt != res)  rev[u] = INF;
    else f[u] += dep[u] - rev[u];
    if (nex[h[u]] == -1 && u != 1) rev[u] = 0, f[u] = dep[u];
}

int main()
{
    n = read();
    memset(h, -1, sizeof(h));
    for (int i = 2; i <= n; i++)
    {
        int a = read(), b = read();
        add(a, b), add(b, a);
    }
    dfs(1, 0);
    printf("%d", f[1]);
}

T2:

　　把点的坐标从二维转为一维的时候没判纵坐标刚好是r的情况，丢了80pts

　　二分答案就行，判断答案可行的时候优先队列优化一下就好

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

#define mp make_pair
#define int long long

typedef pair<int, int> PII;

inline int read()
{
    int 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;
}

const int N = 100010;

int n, m, K;
int tu[N], tc[N];
int dx[6] = { 0,0,-1,-1,1,1 }, dy[6] = { -1,1,0,1,-1,0 };

inline int get(int id, int k)
{
    int x = id / m + 1, y = id % m;
    y == 0 ? y = m : 0;

    int a = x + dx[k], b = y + dy[k];
    if (!a || !b || a > n || b > m) return 0;
    return (a - 1) * m + b;
}

inline int ned(int u, int k, int d, priority_queue<PII>& que)
{
    int v = get(u, k);
    if (!v) return 0;

    int x = 0;
    if (tc[v] < d)
    {
        x = d - tc[v];
        tc[v] = d;
        que.push(mp(tc[v], v));
    }
    return x;
}

inline int check(int x)
{
    priority_queue<PII> que;
    memcpy(tc, tu, sizeof(tu));
    for (int i = 1; i <= n * m; i++) que.push(mp(tc[i], i));

    int res = 0;
    while (!que.empty())
    {
        int u = que.top().second, vd = que.top().first;
        que.pop();

        if (vd ^ tc[u]) continue;

        int d = tc[u] - x;
        res += ned(u, 0, d, que);
        res += ned(u, 1, d, que);
        res += ned(u, 2, d, que);
        res += ned(u, 3, d, que);
        res += ned(u, 4, d, que);
        res += ned(u, 5, d, que);
        
    }
    return res;
}

signed main()
{
    n = read(), m = read(), K = read();

    for (int i = 1, k = 1; i <= n; i++)
    {
        for (int j = 1; j <= m; j++, k++)
        {
            int x = read();
            tu[k] = x;
        }
    }

    int l = 0, r = 1e12;
    while (l < r)
    {
        register int mid = (l + r) >> 1;
        if (check(mid) <= K) r = mid;
        else l = mid + 1;
    }

    printf("%lld", l);
}

T3:

　　就会10pts暴力

　　先类似于数位DP的预处理出C(i,j)，表示当二进制下是1的最高位是i，一共有j个1的方案数。之后对于每个询问，枚举1的数量i。可以发现一个区间按1的数量分组，分为w组，然后把这些组排序后发现区间[a,b]中最多只有边界的两组被拆分，所以把在[a,b]内的区间都加起来，剩下两个单独处理就行了

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

typedef long long LL;

int C[50][50];

inline int read()
{
    int 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;
}

int count(int n, int i) 
{
    int c = 0, ans = 0;
    for (int j = 29; ~j; j--)
    {
        if (n >> j & 1)
        {
            if (i >= c) ans += C[j][i - c];
            ++c;
        }
    }
    return ans + (c == i);
}

LL sum(int n, int i) 
{
    int c = 0, v = 0; 
    LL ans = 0;
    for (int j = 29; ~j; j--)
    {
        if (n >> j & 1)
        {
            if (i >= c) ans += (LL)v * C[j][i - c];
            if (i > c && j > 0) ans += ((1ll << j) - 1) * C[j - 1][i - c - 1];
            ++c, v += 1 << j;
        }
    }
    return ans + (c == i ? n : 0);
}

int get_rank(int n, int i, int k)
{
    int c = count(n, i), l = 1, r = n;
    while (l <= r)
    {
        int m = l + r >> 1;
        if (c - count(m - 1, i) >= k) l = m + 1;
        else r = m - 1;
    }
    return l - 1;
}

int main() 
{
    for (int i = 0; i <= 30; i++)
    {
        C[i][0] = 1;
        for (int j = 1; j <= i; j++) C[i][j] = C[i - 1][j] + C[i - 1][j - 1];
    }

    int T = read();
    while (T--) 
    {
        int L = read(), R = read(), a = read(), b = read();

        int cnt = 0; 
        LL ans = 0;
        for (int i = 30; i; i--)
        {
            int c = count(R, i) - count(L - 1, i);
            if (cnt < a && b <= cnt + c) 
            {
                int r = get_rank(R, i, a - cnt), l = get_rank(R, i, b - cnt);
                ans += sum(r, i) - sum(l - 1, i);
            }
            else if (cnt < a && a <= cnt + c) 
            {
                int r = get_rank(R, i, a - cnt);
                ans += sum(r, i) - sum(L - 1, i);
            }
            else if (a <= cnt && cnt + c < b) ans += sum(R, i) - sum(L - 1, i);
            else if (a <= cnt && cnt < b) 
            {
                int l = get_rank(R, i, b - cnt);
                ans += sum(R, i) - sum(l - 1, i);
            }

            cnt += c;
            if (cnt > b) break;
        }
        printf("%lld\n", ans);
    }
}

T4:

　　10pts暴力

总结：

　　第二题又粗心写炸了，写代码一定得认真点才行