Codeforces Round #481 (Div. 3)

${\mathrm{D.\; Almost\; Arithmetic\; Progression}}^{}$

题解：枚举公差，a[2]的操作有三种情况：a[2]+1,a[2],a[2]-1；a[1]的操作有三种情况：a[1]+1,a[1],a[1]-1。

#pragma warning(disable:4996)
#include<string>
#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
using namespace std;

const int maxn = 100005;
const int INF = 1e9 + 7;

int n;
int a[maxn];

int cal(int d,int last) {
    int cnt = 0;
    for (int i = 3; i <= n; i++) {
        if (a[i] - last == d) {
            last = a[i];
        }
        else if (a[i] - 1 - last == d) {
            cnt++;
            last = a[i] - 1;
        }
        else if (a[i] + 1 - last == d) {
            cnt++;
            last = a[i] + 1;
        }
        else return -1;
    }
    return cnt;
}

int main()
{
    while (cin >> n) {
        for (int i = 1; i <= n; i++) cin >> a[i];
        if (n == 1 || n == 2) cout << "0" << endl;
        else {
            int ans = INF;

            int d1 = a[2] + 1 - a[1];
            int last = a[2] + 1;
            int x = cal(d1, last);
            if (x != -1) ans = min(ans, x + 1);
            
            d1 = a[2] + 2 - a[1];
            last = a[2] + 1;
            x = cal(d1, last);
            if (x != -1) ans = min(ans, x + 2);

            d1 = a[2] - (a[1] + 1);
            last = a[2];
            x = cal(d1, last);
            if (x != -1) ans = min(ans, x + 1);

            d1 = a[2] - (a[1] - 1);
            last = a[2];
            x = cal(d1, last);
            if (x != -1) ans = min(ans, x + 1);

            d1 = a[2] - 1 - (a[1] + 1);
            last = a[2] - 1;
            x = cal(d1, last);
            if (x != -1) ans = min(ans, x + 2);

            d1 = a[2] - 1 - a[1];
            last = a[2] - 1;
            x = cal(d1, last);
            if (x != -1) ans = min(ans, x + 1);

            d1 = a[2] - a[1];
            last = a[2];
            x = cal(d1, last);
            if (x != -1) ans = min(ans, x);
            
            d1 = (a[2] + 1) - (a[1] + 1);
            last = a[2] + 1;
            x = cal(d1, last);
            if (x != -1) ans = min(ans, x + 2);

            d1 = (a[2] - 1) - (a[1] - 1);
            last = a[2] - 1;
            x = cal(d1, last);
            if (x != -1) ans = min(ans, x + 2);

            if (ans == INF) {
                cout << "-1" << endl;
            }
            else cout << ans << endl;
        }
    }
    return 0;
}

E. Bus Video System

$\mathrm{题解：前缀和表示当前这个车站相对于起点（起点车站标为：0）下了多上人或者上了多少人。那么Max和Min分别表示极限情况。Min说明起点至少有多少人(abs(Min))，Max说明起点至多能有多少人（m-Max)。所以答案就是（m-Max)-abs(Min)+1。}$

#pragma warning(disable:4996)
#include<string>
#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
using namespace std;

const int maxn = 1005;

int n, m;

int main()
{
    while (cin >> n >> m) {
        int res = 0, Max = 0, Min = 0;
        for (int i = 1; i <= n; i++) {
            int d;
            cin >> d;
            res += d;
            Max = max(Max, res);
            Min = min(Min, res);
        }
        int ans = m - Max + min(0, Min) + 1;
        if (ans < 0) {
            cout << 0 << endl;
        }
        else {
            cout << ans << endl;
        }
    }
    return 0;
}

 F. Mentors

题解：哈希后求前缀和，可以得到小于当前数的有多少个，然后统计争吵中小于当前数的有多少个。

感受：这题没多大意思，因为给出的争吵关系不具有传递性~~~，如果有传递性就灰常麻烦了，我还多写了并查集 ：( 。

#pragma warning(disable:4996)
#include<map>
#include<string>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

const int maxn = 200005;

int n, m;
int pa[maxn], d[maxn], c[maxn], cnt[maxn], ans[maxn];

map<int, int> p;

int Find(int a) {
    if (a == pa[a]) return a;
    return pa[a] = Find(pa[a]);
}

void Union(int a, int b) {
    int x = Find(a);
    int y = Find(b);
    if (x != y) pa[y] = x;
}

void Inite() {
    p.clear();
    memset(cnt, 0, sizeof(cnt));
    memset(ans, 0, sizeof(ans));
    for (int i = 1; i <= n; i++) pa[i] = i;
}

int main()
{
    while (scanf("%d%d", &n, &m) != EOF) {

        Inite();
        for (int i = 1; i <= n; i++) {
            scanf("%d", &c[i]);
            d[i] = c[i];
        }

        sort(d + 1, d + n + 1);
        int k = unique(d + 1, d + n + 1) - (d + 1);
        for (int i = 1; i <= k; i++) p[d[i]] = i;

        for (int i = 1; i <= m; i++) {
            int u, v;
            scanf("%d%d", &u, &v);
            if (c[u] > c[v]) ans[u]++;
            if (c[u] < c[v]) ans[v]++;
        }

        cnt[0] = 0;
        for (int i = 1; i <= n; i++) cnt[p[c[i]]]++;
        for (int i = 1; i <= k; i++) cnt[i] += cnt[i - 1];

        for (int i = 1; i <= n; i++) {
            cout << cnt[p[c[i]] - 1] - ans[i] << endl;
        }

    }
    return 0;
}