题目大意

给你一个N×M的布，你可以将最多L块布同时剪一刀，问你把他全部剪成1×1的最少要多少刀

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解题思路#1

直接从中间剪，然后dfs求出一个图，然后每次找L个点去跑

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代码#1

#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
#define N 10000
using namespace std;
int k, n, m, w, ans, s[N], to[N][2];
struct node
{int v, s;bool operator < (const node &b) const{return s < b.s;}
};
priority_queue<node>d;
queue<int>dd;
int dfs(int x, int y)
{if (x == 1 && y == 1) return 0;int g = ++w, gg;s[g] = 1;if (x < y) swap(x, y);gg = dfs(x / 2, y);//从中间切s[g] += s[gg];to[g][0] = gg;gg = dfs(x - x / 2, y);s[g] += s[gg];to[g][1] = gg;;return g;
}
void bfs(int x)
{d.push((node){x, s[x]});while(!d.empty()){ans++;int g = 0;while(g < k && !d.empty())//每次找k个点{dd.push(d.top().v);d.pop();g++;}while(!dd.empty()){int h = dd.front();dd.pop();if (to[h][0]) d.push((node){to[h][0], s[to[h][0]]});if (to[h][1]) d.push((node){to[h][1], s[to[h][1]]});}}return;
}
int main()
{scanf("%d%d%d", &k, &n, &m);bfs(dfs(n, m));printf("%d", ans);return 0;
}

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解题思路#2

按上面的剪法，现有的布都能剪，那么能剪就剪即可

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代码#2

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
using namespace std;
int n, m, k, g, ans;
int main()
{scanf("%d%d%d", &k, &n, &m);g = 1;while(g < n * m){g = g + min(g, k);ans++;}printf("%d",ans);return 0;
}