正睿20秋季提高十连测day5

估分：75+23=98

实际：75+23=98

 

T1：

　　思路对了，但少考虑了二分图和不连通图的情况

　　连通图中，k=0，每个点度数必须为偶，k≥1，每个点度数可以全为偶或全为奇，k≥2，每个点度数可以全为偶或全为奇，若图为二分图还可以每条边都是两端端点一个为奇一个为偶。特别判断一条链的情况，k若大于等于边数，也是满足的（因为可以变为单点或空图）。

　　非连通图中，只有一个连通块可以不是一条链，其他连通块必须全部为链或单个点，且k能够把其他为链或单点的连通块全变成空图。之后把不是一条链的连通块当连通图考虑。

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

inline int read() 
{
    int f = 0, x = 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;

int n, m, k;
int u[N], v[N], sz[N], din[N], fa[N], e[N];

int find(int x)
{
    return x == fa[x] ? x : fa[x] = find(fa[x]);
}

int main() 
{
    n = read(), m = read(), k = read();

    for (int i = 1; i <= n; i++)  fa[i] = i, sz[i] = 1;

    for (int i = 1; i <= m; i++) 
    {
        int a = read(), b = read();
        u[i] = a, v[i] = b;
        din[a]++, din[b]++;
        a = find(a), b = find(b);
        if (a != b) fa[a] = b, sz[b] += sz[a], e[b] += e[a];
        e[b]++;
    }

    bool flag = 1, flag1 = 1, flag2 = 1;
    int cnt = 0, cnt2 = 0;
    for (int i = 1; i <= n; i++)
    {
        flag &= !(din[i] & 1);
        if (i == find(i) && sz[i] == 1 && !k)
        {
            printf("No\n");
            return 0;
        }
        if (i == find(i) && sz[i] > 1 && (e[i] >= sz[i] || sz[i] > k)) 
        {
            if (sz[i] == k + 1)
            {
                cnt2++;
                continue;
            }

            cnt++;
            for (int j = 1; j <= m; j++) 
            {
                flag2 &= ((din[u[i]] & 1) ^ (din[v[i]] & 1));
            }
        }
    }
    for (int i = 1; i <= m; i++) flag1 &= !((din[u[i]] + din[v[i]]) & 1);
    if (!cnt && cnt2 == 1 || (cnt == 1 && !cnt2 && (flag || k >= 1 && flag1 || k > 1 && flag2))) printf("Yes\n");
    else  printf("No\n");
}

 

T2：

　　不会，29分的暴力没写对

　　整体二分，用容斥算出被询问的矩形被多少矩形覆盖，然后用树状数组维护，具体看代码

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

typedef long long LL;

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 = 200010;
const int INF = 2147483600;

int n, m;

struct Node
{
    int x, y, id;
    Node() {}
};

bool cmpx(Node a, Node b)
{
    return a.x < b.x;
}

int ans[N + 1], id[N + 1];
int sum[N + 1];

struct Count
{
    Node d[2][N + 1], p[N + 1], q[N + 1];

    inline void add(int q, int id, int x, int y)
    {
        d[q][id].id = id;
        d[q][id].x = x + n + 1;
        d[q][id].y = y + n + 1;
        return;
    }

    int c[N + 1];
    inline void mdf(int x, int d)
    {
        for (; x <= 2 * n + 2; x += (x & (-x))) c[x] += d;
    }

    inline int qry(int x)
    {
        int ret = 0; for (; x; x -= (x & (-x))) ret += c[x]; return ret;
    }

    inline void Move(int l, int r, int ql, int qr, int n)
    {
        int np = 0, nq = 0;
        for (int i = l; i <= r; i++) p[++np] = d[0][i];
        for (int i = ql; i <= qr; i++) q[++nq] = d[1][id[i]];
        sort(p + 1, p + np + 1, cmpx); sort(q + 1, q + nq + 1, cmpx);
        int now = 1;
        for (int i = 1; i <= nq; i++)
        {
            while (now <= np && p[now].x < q[i].x) { mdf(p[now].y, 1); ++now; }
            sum[q[i].id] += n * (now - 1 - qry(q[i].y - 1));
        }
        for (int i = 1; i < now; i++) mdf(p[i].y, -1);
        return;
    }
}L, R, D, U;

int K[N + 1];
int p[N + 1], q[N + 1];

inline void divide(int l, int r, int ql, int qr)
{
    if (ql > qr) return;
    if (l == r)
    {
        for (int i = ql; i <= qr; i++) ans[id[i]] = l;
        return;
    }
    int mid = (l + r) >> 1;
    L.Move(l, mid, ql, qr, 1);
    R.Move(l, mid, ql, qr, 1);
    U.Move(l, mid, ql, qr, 1);
    D.Move(l, mid, ql, qr, 1);

    int np = 0, nq = 0;
    for (int i = ql; i <= qr; i++)
    {
        if (K[id[i]] <= mid - sum[id[i]]) p[++np] = id[i];
        else q[++nq] = id[i];
    }

    for (int i = ql; i <= ql + np - 1; i++) id[i] = p[i - ql + 1];
    for (int i = ql + np; i <= qr; i++) id[i] = q[i - ql - np + 1];
    int md = ql + np - 1;
    L.Move(l, mid, ql, md, -1);
    R.Move(l, mid, ql, md, -1);
    U.Move(l, mid, ql, md, -1);
    D.Move(l, mid, ql, md, -1);
    divide(l, mid, ql, md);
    divide(mid + 1, r, md + 1, qr);
}

int main()
{
    n = read(); m = read();
    for (int i = 1; i <= n; i++)
    {
        int x1 = read(), y1 = read(), x2 = read(), y2 = read();
        L.add(0, i, x2, y2);
        R.add(0, i, -x1, -y1);
        U.add(0, i, -y1, x2);
        D.add(0, i, y2, -x1);
    }
    for (int i = 1; i <= m; i++)
    {
        int x1 = read(), y1 = read(), x2 = read(), y2 = read();
        L.add(1, i, x1, y1);
        R.add(1, i, -x2, -y2);
        U.add(1, i, -y2, x1);
        D.add(1, i, y1, -x2);
        K[i] = read();
    }

    for (int i = 1; i <= m; i++) id[i] = i;
    divide(1, n + 1, 1, m);
    for (int i = 1; i <= m; i++)
    {
        if (ans[i] <= n) printf("%d\n", ans[i]);
        else puts("-1");
    }
}

 

T3:

　　尽力了，真不会。思路看不懂，代码也看不懂。

 

总结:

　　T1虽然想出思路了，但特殊的情况还是没有考虑到，以后还是要想仔细点；容斥不大会用，还要多刷题，关于这种图上的数学期望题做少了，不会，数学期望不太会算，要多做些题。