AtCoder Regular Contest 102 D - All Your Paths are Different Lengths

Posted 2021-01-04 widsom

D - All Your Paths are Different Lengths

思路：

二进制构造

首先找到最大的t，使得2^t <= l

然后我们就能构造一种方法使得正好存在 0 到 2^t - 1 的路径

方法是：对于节点 i 到 i + 1，添加两条边，一条边权值是2^(i-1)，一条边权值是0

对于剩下的2^t 到 l-1的路径，我们考虑倍增地求，每次添加一条节点 v 到 节点 n 的边，边的权值是 X ，新增的路径是X 到 X + 2^(v-1) - 1

第一次的X是 2^t，之后每次倍增X增加 2^v，使得 X + 2^v <= l 

代码：

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pii pair<int, int>
#define piii pair<pii, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//head

vector<piii> ans;
int main() {
    int l, t, n;
    scanf("%d", &l);
    for (int i = 25; ; i--) {
        if((1<<i) <= l) {
            t = i;
            n = t+1;
            break;
        }
    }
    for (int i = 1; i <= t; i++) {
        ans.pb({{i, i+1}, 1<<i-1});
        ans.pb({{i, i+1}, 0});
    }
    int res = l - (1<<t), now = (1<<t);
    for (int i = n-1; i >= 1; i--) {
        if((1<<i-1) <= res) {
            res -= 1<<i-1;
            ans.pb({{i, n}, now});
            now += 1<<i-1;
        }
    }
    printf("%d %d
", n, (int)ans.size());
    for (int i = 0; i < ans.size(); i++) printf("%d %d %d
", ans[i].fi.fi, ans[i].fi.se, ans[i].se);
    return 0;
}