hdu 4671 백업 계획 (구조)

제목 링크: hdu 4671 Backup Plan
코드
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>

using namespace std;
typedef pair<int,int> pii;
const int maxn = 105;

int N, K, G[maxn][maxn], C[maxn];
vector<pii> s;

void solve(int l, int r, int k) {
    s.clear();
    for (int i = 0; i < N; i++) if (i != k)
        s.push_back(make_pair(C[i], i));
    sort(s.begin(), s.end());

    for (int i = l; i <= r; i++) {
        int t = (i - l) % (N-1);
        G[i][1] = s[t].second;
    }
}

void draw(int* x) {
    int p = 2;
    for (int i = 0; i < N; i++) {
        if (i == x[0] || i == x[1]) continue;
        x[p++] = i;
    }
}

int main () {
    while (scanf("%d%d", &N, &K) == 2) {
        memset(C, 0, sizeof(C));
        int t = K / N, f = K % N, p = 0;

        for (int i = 0; i < f; i++) {
            for (int j = 0; j < t + 1; j++)
                G[p++][0] = i;
            C[i] = t + 1;
        }
        for (int i = f; i < N; i++) {
            for (int j = 0; j < t; j++)
                G[p++][0] = i;
            C[i] = t;
        }

        p = 0;
        for (int i = 0; i < K; i++) {
            if (G[i][0] != G[p][0]) {
                solve(p, i - 1, G[p][0]);
                p = i;
            }
        }
        solve(p, K-1, G[p][0]);

        for (int i = 0; i < K; i++) {
            draw(G[i]);
            for (int j = 0; j < N; j++)
                printf("%d%c", G[i][j] + 1, j == N-1 ? '
'
: ' '); } } return 0; }

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