[Bzoj3193] [JLOI2013] Terrain generate permutations _ _ greedy

[JLOI2013] terrain generation

Topic links : https://www.lydsy.com/JudgeOnline/problem.php?id=3193

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Problem solution :

This arrangement required a total of the title, one conventional approach is that all the elements are arranged according to a certain manner and then inserted inside a a.

The problem is sorted in descending order like this has no effect on the elements behind.

There are a number of elements equal to how we get?

If the label is a sequence of words is blind $ jb $ row on the line.

If the contour is the sequence, so long as the $ A $ in accordance with the second keyword like.

Code :

#include <bits/stdc++.h>

#define N 1010 

using namespace std;

const int mod = 2011 ;

char *p1, *p2, buf[100000];

#define nc() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? EOF : *p1 ++ )

int rd() {
    int x = 0, f = 1;
    char c = nc();
    while (c < 48) {
        if (c == '-')
            f = -1;
        c = nc();
    }
    while (c > 47) {
        x = (((x << 2) + x) << 1) + (c ^ 48), c = nc();
    }
    return x * f;
}

struct Node {
    int h, rk;
}a[N];

int dp[N];

inline bool cmp(const Node &a, const Node &b) {
    return a.h == b.h ? a.rk < b.rk : a.h > b.h;
}

int main() {
    int n = rd();
    for (int i = 1; i <= n; i ++ ) {
        a[i].h = rd(), a[i].rk = rd() - 1;
    }
    int ans1 = 1, ans2 = 1;
    sort(a + 1, a + n + 1, cmp);
    for (int i = 1; i <= n; i ++ ) {
        int dic = i;
        while (dic <= n && a[dic].h == a[i].h) {
            dic ++ ;
        }
        dic -- ;
        memset(dp, 0, sizeof dp);
        dp[0] = 1;
        for (int j = i; j <= dic; j ++ ) {
            ans1 = ans1 * (min(i, a[j].rk + 1) + j - i) % mod;
            for (int k = 1; k <= min(a[j].rk, i - 1); k ++ ) {
                dp[k] = (dp[k - 1] + dp[k]) % mod;
            }
        }
        int sum = 0;
        for (int j = 0; j <= min(a[dic].rk, i - 1); j ++ ) {
            (sum += dp[j]) %= mod;
        }
        (ans2 *= sum) %= mod;
        i = dic;
    }
    cout << ans1 << ' ' << ans2 << endl ;
    return 0;
}

Summary : Remember, to consider the issue when the current time stamp that is easy to overlook the variables taken into account, even though most of the time with less.