Codeforces Round #413(Div. 1 + Div. 2)

C. Fountains

首先特判买一个C一个D的情况。
考虑买两个C(或D)，先按照价格排个序，枚举一个位置，则另一个可行位置构成一个区间。然后用ST表区间最值即可。

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 100005;
struct ft {
    int b, c;
    friend bool operator < (const ft &a, const ft &b)
    { return a.c < b.c; }
} coin[MAXN], dm[MAXN];
int top1 = 0, top2 = 0;
int max_c[MAXN], max_d[MAXN];
int n, c, d;
int f[MAXN][25], lg[MAXN];
int max_val(int i, int j)
{
    if (i > j) return 0;
    int k = lg[j-i+1];
    return max(f[i][k], f[j-(1<<k)+1][k]);
}
int max_val_without(int i, int j)
{
    if (i < j) return max_val(1, i);
    return max(max_val(1, j-1), max_val(j+1, i));
}
int solve(ft a[MAXN], int top, int M)
{
    sort(a+1, a+top+1);
    memset(f, 0, sizeof f);
    for (int i = 1; i <= top; i++) f[i][0] = a[i].b;
    for (int j = 1; j <= 20; j++) 
        for (int i = 1; i <= top; i++) {
            f[i][j] = f[i][j-1];
            if (i+(1<<(j-1)) <= top) f[i][j] = max(f[i][j], f[i+(1<<(j-1))][j-1]);
        }
    int Rt = top, ans = 0;
    for (int i = 1; i <= top; i++) {
        if (a[i].c > M) break;
        while (Rt && (a[Rt].c+a[i].c > M || Rt == i)) Rt--;
        if (Rt == 0) break;
        ans = max(ans, max_val_without(Rt, i)+a[i].b);
    } 
    //cout << ans << endl;
    return ans;
}
int main()
{
    scanf("%d %d %d", &n, &c, &d);
    int j = 1, t = 0;
    for (int i = 1; i <= n; i++) {
        while (j*2 <= i) j <<= 1, t++;
        lg[i] = t;
        //cout << lg[i] << " ";
    }
    //cout << endl;
    for (int i = 1; i <= n; i++) {
        int a, b; char ch;
        scanf("%d %d %c", &a, &b, &ch);
        // cout << a << " " << b << "," << c << endl;
        if (ch == 'C') coin[++top1].b = a, coin[top1].c = b, max_c[b] = max(max_c[b], a);
        else dm[++top2].b = a, dm[top2].c = b, max_d[b] = max(max_d[b], a);
    }
    max_d[0] = 0, max_c[0] = 0;
    for (int i = 1; i <= c; i++) max_c[i] = max(max_c[i], max_c[i-1]);
    for (int i = 1; i <= d; i++) max_d[i] = max(max_d[i], max_d[i-1]);
    // case1 
    int ans = 0;
    if (max_d[d] != 0 && max_c[c] != 0)
        ans = max(ans, max_d[d]+max_c[c]);
    // case2
    ans = max(ans, max(solve(coin, top1, c), solve(dm, top2, d)));
    cout << ans << endl;
    return 0;
}

D. Field expansion

大力爆搜..CF机子也真是快...

#include <bits/stdc++.h>
using namespace std;
long long a, b, h, w, n;
long long ai[100005];
long long dat[100005], top = 0;
int tms[100005];
int dfs(int nd, long long h, long long w)
{
    //cout << dat[nd] << ' ' << h << ' ' << w << endl;
    if ((h >= a && w >= b) ||(h >= b && w >= a)) return 0;
    if (nd == 0) return 2333333;
    long long th = 1, tw = 1;
    int ans = 2333333;
    for (int i = 0; i <= tms[dat[nd]]; i++) {
        tw = 1;
        for (int j = 0; j+i <= tms[dat[nd]]; j++) {
            ans = min(ans, dfs(nd-1, h*th, w*tw)+i+j);
            if (w*tw >= 100000) break;
            tw *= dat[nd];
        }
        if (h*th >= 100000) break;
        th *= dat[nd];
    }
    return ans;
}
int main()
{
    scanf("%I64d%I64d%I64d%I64d%I64d", &a, &b, &h, &w, &n);
    for (int i = 1; i <= n; i++) scanf("%I64d", &ai[i]);
    sort(ai+1, ai+n+1); 
    for (int i = n; i >= max(1ll, n-40); i--) { 
        tms[ai[i]]++;
        if (dat[top] != ai[i])
            dat[++top] = ai[i];
        // cout << i << endl;
    }
    int ans = dfs(top, h, w);
    if (ans < 2333333)
    cout << dfs(top, h, w) << endl;
    else puts("-1");
    return 0;
}