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牛客多校4 H

Wall Builder II

题目链接

题目大意

给出\(n\)种砖块，砖块高度为\(1\)，宽度为\(1-n\)，面积为\(i\)的砖块数量有\(n-i+1\)个，问砌成周长最小的矩形的方案并输出

题解

简单贪心，要使砌成的矩形周长最小，不难想到应该越像个正方形越小，接下来考虑怎么填，我们发现长度越小的灵活度越高，因此优先考虑放大的，一块一块填就行

代码

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#include<bits/stdc++.h>
#define int long long
using namespace std;
const int N = 105;
int cnt[N];
struct node_ {
    int x1, y1, x2, y2;
};

vector<node_> res[N];
int tmp[N * N];
int find(int l, int r, int x) {
    while(l < r) {
        int mid = (l + r + 1) >> 1;
        if(tmp[mid] > x) r = mid - 1;
        else l = mid;
    }
    return r;
}
signed main() {
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(0);
    int t;
    cin >> t;
    while(t -- ) {
        memset(cnt, 0, sizeof(cnt));
        //memset(tmp, 0, sizeof(tmp));
        int n, sumv = 0;
        cin >> n;
        int kk = 0;
        for(int i = 1, j = n; i <= n; i ++, j -- ) {
            sumv += i * j;
            cnt[i] = j;
            res[i].clear();
            for(int k = 1; k <= j; k ++ ) tmp[++ kk] = i;
        }
//         for(int i = 1; i <= kk; i ++ ) cout << tmp[i] << " ";
//         cout << '\n';
        int maxx = 1;
        for(int i = 1; i <= sumv / i; i ++ ) {
            if(sumv % i == 0) {
                maxx = max(maxx, i);
            }
        }
//         sort(tmp + 1, tmp + kk + 1);
        int maxlen = sumv / maxx;
        for(int i = 1; i <= maxx; i ++ ) {
            int nowlen = 0;
            int pos = n;
            while(nowlen != maxlen) {
                for(int j = n; j >= 1; j -- ) {
                    if(cnt[j] && j <= maxlen - nowlen) {
                        pos = j;
                        break;
                    }
                }
                nowlen += pos;
                cnt[pos] -- ;
                res[pos].push_back({nowlen - pos, i - 1, nowlen, i});
                if(nowlen == maxlen) break;
            }
        }
        cout << 2 * (maxlen + maxx) << '\n';
        for(int i = 1; i <= n; i ++ ) {
            for(auto k : res[i]) {
                cout << k.x1 << " " << k.y1 << " " << k.x2 << " " << k.y2 << '\n';
            }
        }
    }
    return 0;
}