題目大意

給你一個N×M的布，你可以將最多L塊布同時剪一刀，問你把他全部剪成1×1的最少要多少刀

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解題思路#1

直接從中間剪，然后dfs求出一個圖，然后每次找L個點去跑

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代碼#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);//從中間切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個點{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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解題思路#2

按上面的剪法，現(xiàn)有的布都能剪，那么能剪就剪即可

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代碼#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; }

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