2019-2020 ICPC North-Western Russia Regional Contest

最近事务繁多，很久没写文章。正好趁刚换新队友的第一场训练，记录一下。

题目中规中矩。

---

A：

solver：lzh

完全没看，好像是个傻逼题。

#include<bits/stdc++.h>
using namespace std;
int main() {
    int a, b, n; scanf("%d%d%d", &a, &b, &n);
    int c = b, d = a, ans = 0, cur = 1;
    while (c != n || d != n) {
        if (!cur) {
            c += b - a;
            if (c > n)c = n;
        } else {
            d += b - a;
            if (d > n)d = n;
        }
        cur ^= 1;
        ans++;
    }
    printf("%d
", ans);
}

E：

solver：zyh、lzh

给定一棵树，树上有若干个重要点。问是否存在一个点到所有重要点距离相等。

树的直径瞎搞。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
vector<int>v[200010];
int mark[200010], ismark[200010];
int st[200010], num = 0, ans = 0, ansn;
int dep[200010];
void dfs1(int x, int pre, int d) {
    st[++num] = x;
    if (ismark[x] && ans < d)ans = d, ansn = st[num + 1 >> 1];
    for (auto i : v[x])
        if (i != pre)
            dfs1(i, x, d + 1);
    num--;
}
void dfs2(int x, int pre, int d) {
    dep[x] = d;
    for (auto i : v[x])
        if (i != pre)
            dfs2(i, x, d + 1);
}
int main() {
    int n, m; scanf("%d%d", &n, &m);
    for (int i = 2; i <= n; i++) {
        int x, y; scanf("%d%d", &x, &y);
        v[x].push_back(y);
        v[y].push_back(x);
    }
    for (int i = 1; i <= m; i++)scanf("%d", &mark[i]), ismark[mark[i]]++;
    if (m == 1) {
        printf("YES
1
");
        return 0;
    }
    dfs1(mark[1], -1, 0);
    if (ans & 1) {
        printf("NO
"); return 0;
    }
    dfs2(ansn, -1, 1);
    for (int i = 2; i <= m; i++)
        if (dep[mark[i - 1]] != dep[mark[i]]) {
            printf("NO
");
            return 0;
        }
    printf("YES
%d
", ansn);
}

H：

solver：czq

给定n个数，要求把这n个数分成连续的尽量少的块，并且每块和<=t。

由于有多个询问，故可以预处理出每个询问能装多少个数。然后每个询问直接模拟统计答案。

/* basic header */
#include <bits/stdc++.h>
/* define */
#define ll long long
#define dou double
#define pb emplace_back
#define mp make_pair
#define sot(a,b) sort(a+1,a+1+b)
#define rep1(i,a,b) for(int i=a;i<=b;++i)
#define rep0(i,a,b) for(int i=a;i<b;++i)
#define eps 1e-8
#define int_inf 0x3f3f3f3f
#define ll_inf 0x7f7f7f7f7f7f7f7f
#define lson (curpos<<1)
#define rson (curpos<<1|1)
/* namespace */
using namespace std;
/* header end */

const int maxn = 1e6 + 10;
int n, a[maxn], pos[maxn], cnt[maxn], preSum[maxn], maxx;

int solve(int x) {
    if (x < maxx) return -int_inf;
    if (cnt[x]) return cnt[x];
    int res = 0, i = 0;
    while (i < n) {
        res += x;
        res = min(res, 1000000);
        i = pos[res];
        res = preSum[i];
        cnt[x]++;
    }
    return cnt[x];
}

int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        maxx = max(maxx, a[i]);
        preSum[i] = preSum[i - 1] + a[i];
        for (int j = preSum[i - 1]; j < preSum[i]; j++) pos[j] = i - 1;
    }
    for (int i = preSum[n]; i <= 1000000; i++) pos[i] = n;
    int q; scanf("%d", &q);
    while (q--) {
        int x; scanf("%d", &x);
        int ret = solve(x);
        if (ret != -int_inf) printf("%d
", solve(x));
        else puts("Impossible");
    }
    return 0;
}

I：

solver：czq

给定三维空间内的n个点，要求构造一个尽量小的四棱锥(底面两边与坐标轴平行，且锥面与底面夹角为45度)，使得该四棱锥能包含n个点。

因为锥面与底面夹角为45度，故可以拆开分别考察xOz和yOz两个平面，构造出两个等腰直角三角形之后再合并答案。

/* basic header */
#include <bits/stdc++.h>
/* define */
#define ll long long
#define dou double
#define pb emplace_back
#define mp make_pair
#define sot(p,b) sort(p+1,p+1+b)
#define rep1(i,p,b) for(int i=p;i<=b;++i)
#define rep0(i,p,b) for(int i=p;i<b;++i)
#define eps 1e-8
#define int_inf 0x3f3f3f3f
#define ll_inf 0x7f7f7f7f7f7f7f7f
#define lson (curpos<<1)
#define rson (curpos<<1|1)
/* namespace */
using namespace std;
/* header end */

const int maxn = 1e3 + 10;
struct Point {
    int x, y, z;
} p[maxn];
int n, l1 = INT_MAX, l2 = INT_MAX, r1 = -INT_MAX, r2 = -INT_MAX;

int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d%d%d", &p[i].x, &p[i].y, &p[i].z);
    for (int i = 1; i <= n; i++) {
        l1 = min(p[i].x - p[i].z, l1); l2 = min(p[i].y - p[i].z, l2);
        r1 = max(p[i].x + p[i].z, r1); r2 = max(p[i].y + p[i].z, r2);
    }
    int currx = l1 + r1 >> 1, curry = l2 + r2 >> 1;
    int h1 = max(currx - l1, curry - l2), h2 = max(r1 - currx, r2 - curry);
    printf("%d %d %d
", currx, curry, max(h1, h2));
    return 0;
}

J：

solver：czq

显然倒着构造会好构造很多。对于点对(i,j)，计算所有的k(i<k<j)对a[i][j]的贡献即可。如果不一样，说明点i与点j之间存在边。

/* basic header */
#include <bits/stdc++.h>
/* define */
#define ll long long
#define dou double
#define pb emplace_back
#define mp make_pair
#define sot(ans,b) sort(ans+1,ans+1+b)
#define rep1(i,ans,b) for(int i=ans;i<=b;++i)
#define rep0(i,ans,b) for(int i=ans;i<b;++i)
#define eps 1e-8
#define int_inf 0x3f3f3f3f
#define ll_inf 0x7f7f7f7f7f7f7f7f
#define lson (curpos<<1)
#define rson (curpos<<1|1)
/* namespace */
using namespace std;
/* header end */

const int maxn = 510;
int n, a[maxn][maxn], ans[maxn][maxn];

int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            scanf("%1d", &a[i][j]);
    for (int i = n - 1; i >= 1; i--)
        for (int j = i; j <= n; j++) {
            int contribute = 0;
            for (int k = i + 1; k < j; k++) contribute = (contribute + ans[i][k] * a[k][j]) % 10;
            if (contribute != a[i][j]) ans[i][j] = 1;
        }
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) printf("%d", ans[i][j]);
        puts("");
    }
    return 0;
}

K：

solver：lzh、czq

细节贪心题，枚举A的最大可能上下边界，得出可行的左右边界，再分割剩下的区域。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, char> pic;
typedef pair<int, int> pii;
char s[1010][1010], ans[1010][1010];
pii p[30];
vector<pic> v[1010];
int pt[1010];
void solve(pii l, pii r) {
    for (int i = 1; i <= 1000; i++)v[i].clear(), pt[i] = 0;
    for (int i = 2; i <= 26; i++)
        if (l.first <= p[i].first && p[i].first <= r.first &&
                l.second <= p[i].second && p[i].second <= r.second)
            v[p[i].first].push_back(make_pair(p[i].second, i + 'a' - 1));

    for (int i = l.first; i <= r.first; i++)
        sort(v[i].begin(), v[i].end());

    for (int i = l.first; i <= r.first; i++)
        if (v[i].size() == 0 && pt[i - 1]) {
            for (int j = l.second; j <= r.second; j++)ans[i][j] = ans[i - 1][j];
            pt[i] = 1;
        } else if (v[i].size() == 1) {
            for (int j = l.second; j <= r.second; j++)ans[i][j] = v[i][0].second;
            pt[i] = 1;
        } else if (v[i].size() > 1) {
            pt[i] = 1;
            int pre = l.second;
            for (auto j : v[i]) {
                for (; pre <= j.first; pre++)ans[i][pre] = j.second;
            }

            for (; pre <= r.second; pre++)ans[i][pre] = ans[i][pre - 1];
        }

    for (int i = r.first; i >= l.first; i--)
        if (!pt[i])
            for (int j = l.second; j <= r.second; j++)
                ans[i][j] = ans[i + 1][j];
}
int main() {
    int n, m; scanf("%d%d", &n, &m);
    pii A;
    for (int i = 1; i <= n; i++) {
        scanf("%s", s[i] + 1);
        for (int j = 1; j <= m; j++)
            if (s[i][j] != '.') {
                if (s[i][j] != 'A') {
                    p[s[i][j] - 'A' + 1] = make_pair(i, j);
                    v[i].push_back(make_pair(j, s[i][j] + 'a' - 'A'));
                } else A = make_pair(i, j);
            }
    }

    int tmp = 0;
    pii L, R;
    for (int i = 1; i <= n; i++) {
        priority_queue<int>l;
        priority_queue<int, vector<int>, greater<int>>r;
        l.push(0);
        r.push(m + 1);
        for (int j = i; j <= n; j++) {
            int gg = 0;
            for (int k = 2; k <= 26; k++)
                if (i <= p[k].first && p[k].first <= j) {
                    if (p[k].second == A.second) {
                        gg++; break;
                    } else if (p[k].second < A.second)l.push(p[k].second);
                    else r.push(p[k].second);
                }
            if (gg)break;
            if (i <= A.first && A.first <= j && l.top() < A.second && A.second < r.top()
                    && (j - i + 1) * (r.top() - l.top() - 1) > tmp) {
                tmp = (j - i + 1) * (r.top() - l.top() - 1);
                L = make_pair(i, l.top() + 1);
                R = make_pair(j, r.top() - 1);
            }
        }
    }
    if (L.first != 1)solve(make_pair(1, 1), make_pair(L.first - 1, m));
    if (R.first != n)solve(make_pair(R.first + 1, 1), make_pair(n, m));
    if (L.second != 1)solve(make_pair(L.first, 1), make_pair(R.first, L.second - 1));
    if (R.second != m)solve(make_pair(L.first, R.second + 1), make_pair(R.first, m));

    for (int i = 2; i <= 26; i++)ans[p[i].first][p[i].second] = 'A' + i - 1;
    for (int i = L.first; i <= R.first; i++)
        for (int j = L.second; j <= R.second; j++)
            ans[i][j] = 'a';
    ans[A.first][A.second] = 'A';

    for (int i = 1; i <= n; i++, printf("
"))
        for (int j = 1; j <= m; j++)
            printf("%c", ans[i][j]);
}

M：

solver：zyh

没看，签到。

#include <iostream>
#include <algorithm>
#include <map>
using namespace std;
int a[10001];
map<int, int> M;
int main() {
    int T;
    long long ans = 0;
    scanf("%d", &T);
    while (T--) {
        int n;
        M.clear();
        ans = 0;
        scanf("%d", &n);
        for (int i = 0; i < n; ++i) {
            scanf("%d", &a[i]);
            M[a[i]]++;
        }
        for (int j = 0; j < n; ++j) {
            M[a[j]]--;
            if (M[a[j]] == 0) M.erase(a[j]);
            for (int i = 0; i < j; ++i) {
                int k = 2 * a[j] - a[i];

                if (M.find(k) != M.end()) ans += M[k];
            }
        }
        printf("%lld
", ans);
    }
}