第二周 3.6-3.12

3.6

CF 627 A XOR Equation

a + b = a ^ b + (a & b << 1)

注意非法情况和0。

#include <iostream>
#include <cstdio>
using namespace std;
typedef long long LL;

int main(void)
{
    LL s, x, t, ans = 1LL;
    cin >> s >> x;
    t = s == x ? 2LL : 0LL;
    s -= x;
    if(s < 0LL || s % 2LL) {puts("0"); return 0;}
    s /= 2LL;
    while(x || s)
    {
        if((x % 2LL) && (s % 2LL)) {puts("0"); return 0;}
        if((x % 2LL) && !(s % 2LL)) ans <<= 1LL;
        x /= 2LL, s /= 2LL;
    }
    cout << ans - t << endl;
    return 0;
}

CF 627 B Factory Repairs

两个bit随意搞。

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn = 2e5 + 10;
typedef long long LL;
LL c[2][maxn];

int lowbit(int s)
{
    return s & (-s);
}

void modify(int op, int i, LL x)
{
    while(i < maxn) c[op][i] += x, i += lowbit(i);
    return;
}

LL query(int op, int i)
{
    LL ret = 0LL;
    while(i > 0) ret += c[op][i], i -= lowbit(i);
    return ret;
}

int main(void)
{
    int n, k, a, b, q;
    scanf("%d %d %d %d %d", &n, &k, &a, &b, &q);
    while(q--)
    {
        int op, x, y;
        scanf("%d", &op);
        if(op == 1)
        {
            scanf("%d %d", &x, &y);
            modify(0, x, min(a - query(0, x) + query(0, x - 1), (LL)y));
            modify(1, x, min(b - query(1, x) + query(1, x - 1), (LL)y));
        }
        else
        {
            scanf("%d", &x);
            printf("%I64d
", query(1, x - 1) + query(0, n) - query(0, x + k - 1));
        }
    }
    return 0;
}

CF 627 C Package Delivery

不是很懂贪心。

想起多校某题还不会QAQ。

#include <iostream>
#include <cstdio>
#include <queue>
#include <algorithm>
using namespace std;
typedef long long LL;
const int maxn = 2e5 + 10;

struct gas
{
    int id, x, p;
    friend bool operator < (gas A, gas B)
    {
        return A.p > B.p;
    }
}g[maxn];

bool cmp (gas A, gas B)
{
    return A.x < B.x;
}

int main(void)
{
    int n, d, m;
    scanf("%d %d %d", &d, &n, &m);
    for(int i = 0; i < m; i++) scanf("%d %d", &g[i].x, &g[i].p);
    sort(g, g + m, cmp);
    LL ans = 0LL;
    priority_queue<gas> pq;
    int t = 0, now = n;
    while(now < d)
    {
        while(t < m && g[t].x <= now) pq.push(g[t++]);
        while(!pq.empty() && pq.top().x <= now - n) pq.pop();
        if(pq.empty()) return puts("-1");
        int x = pq.top().x, p = pq.top().p;
        int nxt = min(d, x + n);
        if(t < m && nxt >= g[t].x) nxt = g[t].x;
        ans += (LL) p * (nxt - now);
        now = nxt;
    }
    printf("%I64d
", ans);
    return 0;
}

3.7

CF 627 D Preorder Test

卡题了……崩溃。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 2e5 + 10;
int n, k, a[maxn];

//tree
int cnt, h[maxn];
struct edge
{
    int fm, to, pre;
} e[maxn<<1];

void add(int from, int to)
{
    cnt++;
    e[cnt].pre = h[from];
    e[cnt].fm = from;
    e[cnt].to = to;
    h[from] = cnt;
}

//dp
int sz[maxn], num[maxn];
void dfs1(int mid, int v, int f)
{
    sz[v] = 1;
    num[v] = a[v] >= mid ? 1 : 0;
    for(int i = h[v]; i; i = e[i].pre)
    {
        int to = e[i].to;
        if(to == f) continue;
        dfs1(mid, to, v);
        sz[v] += sz[to];
        num[v] += num[to];
    }
}

int ff[maxn], m[2][maxn], pre[maxn];
void dfs2(int mid, int v, int f)
{
    if(!f) ff[v] = a[v] >= mid ? 1 : 0;
    pre[v] = m[0][v] = m[1][v] = 0;
    for(int i = h[v]; i; i = e[i].pre)
    {
        int to = e[i].to;
        if(to == f) continue;
        if(num[v] - num[to] == sz[v] - sz[to] && ff[v]) ff[to] = 1;
        else ff[to] = 0;
        dfs2(mid, to, v);
        if(a[to] < mid) continue;
        if(sz[to] == num[to]) pre[v] += sz[to];
        else
        {
            if(pre[to] > m[0][v]) m[1][v] = m[0][v], m[0][v] = pre[to];
            else if(pre[to] > m[1][v]) m[1][v] = pre[to];
        }
    }
    if(a[v] < mid) pre[v] = 0;
    else pre[v] += m[0][v] + 1;
}

bool ok(int mid)
{
    dfs1(mid, 1, 0);
    dfs2(mid, 1, 0);
    for(int i = 1; i <= n; i++)
    {
        if(a[i] < mid) continue;
        int tmp = pre[i] + m[1][i];
        if(ff[i]) tmp += n - sz[i];
        if(tmp >= k) return true;
    }
    return false;
}

int main(void)
{
    scanf("%d %d", &n, &k);
    for(int i = 1; i <= n; i++) scanf("%d", a + i);
    for(int i = 1; i < n; i++)
    {
        int u, v;
        scanf("%d %d", &u, &v);
        add(u, v), add(v, u);
    }
    int l = 1, r = 1e6, mid;
    while(l < r)
    {
        mid = r - (r - l) / 2;
        if(ok(mid)) l = mid;
        else r = mid - 1;
    }
    printf("%d
", l);
    return 0;
}

3.8

什么都没干。

3.9 

CF 627 E Orchestra

感觉很少写链表。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;
int r, c, n, k, id[3333];

struct vio
{
    int id, x, y;
}v[3333], cpy[3333];

bool cmp1(vio A, vio B)
{
    return A.x < B.x;
}

bool cmp2(vio A, vio B)
{
    return A.y < B.y;
}

struct node
{
    int y;
    int pre, nxt;
}l[3333];

LL cal(int i, int k)
{
    LL ret = 0LL;
    int p1 = i, p2 = i, t = 0;
    while(l[p1].pre && t < k - 1) p1 = l[p1].pre, t++;
    while(l[p2].nxt && t < k - 1) p2 = l[p2].nxt, t++;
    if(t < k - 1) return ret;
    while(1)
    {
        if(l[p2].nxt) ret += (LL) (l[p1].y - l[l[p1].pre].y) * (l[l[p2].nxt].y - l[p2].y);
        else ret += (LL) (l[p1].y - l[l[p1].pre].y) * (c + 1 - l[p2].y);
        if(p1 == i || !l[p2].nxt) break;
        p1 = l[p1].nxt;
        p2 = l[p2].nxt;
    }
    return ret;
}

void del(int i)
{
    if(l[i].pre) l[l[i].pre].nxt = l[i].nxt;
    if(l[i].nxt) l[l[i].nxt].pre = l[i].pre;
}

int main(void)
{
    scanf("%d %d %d %d", &r, &c, &n, &k);
    for(int i = 0; i < n; i++) scanf("%d %d", &v[i].x, &v[i].y);
    sort(v, v + n, cmp1);
    for(int i = 0; i < n; i++) v[i].id = i;
    memcpy(cpy, v, sizeof(cpy));
    LL ans = 0LL;
    for(int x1 = r; x1; x1--)
    {
        int cnt = 0;
        LL cur = 0LL;
        for(int i = 0; i < n; i++)
        {
            if(v[i].x < x1) continue;
            sort(cpy + i, cpy + n, cmp2);
            for(int j = i; j < n; j++)
            {
                cnt++;
                l[cnt].y = cpy[j].y;
                l[cnt].pre = cnt - 1;
                if(cnt - 1) l[cnt-1].nxt = cnt;
                l[cnt].nxt = 0;
                id[cpy[j].id] = cnt;
            }
            break;
        }
        for(int i = 1; i + k - 1 <= cnt; i++)
            cur += (LL) (l[i].y - l[i-1].y) * (c - l[i+k-1].y + 1);
        int p = n - 1;
        for(int x2 = r; x2 >= x1; x2--)
        {
            ans += cur;
            while(p >= 0 && v[p].x == x2)
            {
                cur -= cal(id[p],k);
                del(id[p]);
                p--;
            }
        }
    }
    printf("%I64d
", ans);
    return 0;
}

3.10-3.12

什么都没干。