Wannafly camp Day1 E Growth

题意

　　二维坐标点(a, b)，初始为0，每个时刻可以让a+1或b+1,。当a>=x_i且b>=y_i时，之后的每个时刻可以获得z_i分。

　　问m时刻后的最大分数。

　　n <= 1000, m <= 2e9, x,y <= 1e9, z <= 1e6

题解

　　考虑到有影响的x和y值只有给定的那些值，所以将x和y分别排序，用t[i][j]表示当a = x_i, b = y_i时能获得的奖励（对于有多组相同的x和y，对z求和），用v[i][j]表示当a = x_i, b = y_i时接下来每天能获得的奖励，即t[i][j]的二维前缀和。

　　考虑DP，f[i][j]可以从i - 1, j和i, j - 1转移过来，方程为：f[i][j] = max(f[i - 1][j] + (x[i] - x[i - 1] - 1) * v[i - 1][j] + v[i][j], f[i][j - 1] + (y[j] - y[j - 1] - 1) * v[i][j - 1] + v[i][j])

　　最后注意扫描找值时要加上剩下天数的贡献。

#include <bits/stdc++.h>
using namespace std;

int x[1010], y[1010], xx[1010], yy[1010], z[1010];

map<pair<int, int > , int> h;

long long t[1010][1010], v[1010][1010];
__int128 f[1010][1010];

void print(__int128 x)
{
    if (x>9) print(x/10);
    putchar('0'+x%10);
}

int main()
{
    int n, m;
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; ++i)
    {
        scanf("%d%d%d", &xx[i], &yy[i], &z[i]);
        if (h[{xx[i], yy[i]}])
            h[{xx[i], yy[i]}] += z[i];
        else
            h[{xx[i], yy[i]}] = z[i];
    }
    sort(xx + 1, xx + n + 1);
    sort(yy + 1, yy + n + 1);
    int X(0), Y(0);
    x[++X] = xx[1];
    y[++Y] = yy[1];
    for (int i = 2; i <= n; ++i)
    {
        if (xx[i] != xx[i - 1])
            x[++X] = xx[i];
        if (yy[i] != yy[i - 1])
            y[++Y] = yy[i];
    }
    for (int i = 1; i <= X; ++i)
        for (int j = 1; j <= Y; ++j)
            t[i][j] = h[{x[i], y[j]}];
    for (int i = 1; i <= X; ++i)
        for (int j = 1; j <= Y; ++j)
            v[i][j] = v[i - 1][j] + v[i][j - 1] - v[i - 1][j - 1] + t[i][j];
    for (int i = 1; i <= X; ++i)
        for (int j = 1; j <= Y; ++j)
            f[i][j] = max(f[i - 1][j] + (__int128)(x[i] - x[i - 1] - 1) * v[i - 1][j] + v[i][j], f[i][j - 1] + (__int128)(y[j] - y[j - 1] - 1) * v[i][j - 1] + v[i][j]);
    __int128 ans(0);
    for (int i = 1; i <= X; ++i)
        for (int j = 1; j <= Y; ++j)
            if (x[i] + y[j] <= m)
                ans = max(ans, f[i][j] + v[i][j] * (m - x[i] - y[j]));
    print(ans);
    putchar('
');

    return 0;
}