[LeetCode 887. 鸡蛋掉落] DP+二分

1. 题目链接

2. 题目描述##

题意描述

3. 解题思路##

首先,令$dp[i][j+1]$表示有$i$个鸡蛋,建筑物为 $j$层时的答案. 那么, 我们容易想到下面的DP的状态转移方程:
$$dp[i][j]=\mathrm{min}{\mathrm{max}(dp[i-1][j-k],dp[i][k])+1}$$
其中, $k\in [0, j]$.
此时, DP的复杂度是$O(KN^2)$. 打表或者直接分析转移方程,可以发现, $dp[i][j]>=dp[i][j-1], dp[i][j]<=dp[i-1][j]$.
那么, 对于给定$i,j$, $dp[i-1][j-k]$随$k$单调递减,$dp[i-1][k]$随$k$单调递增.那么可以二分找到使得 $\mathrm{max}(dp[i-1][j-k],dp[i][k])+1}$最小的$k$.
_{感觉理论上三分应该也是可以的,但是不知道为什么三分就是不太对…}~

4. 参考代码##

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;

const long double eps = 1e-8;
const int inf = 0x3f3f3f3f;
const ll infl = 0x3f3f3f3f3f3f3f3f;

int dp[105][10005];

class Solution {
public:
    int superEggDrop(int K, int N) {
        memset(dp, 0x3f, sizeof(dp));
        for(int i = 0; i <= K; ++i) dp[i][1] = 0;
        for(int j = 1; j <= N + 1; ++j) dp[1][j] = j - 1;
        for(int i = 2; i <= K; ++i) {
            for(int j = 2; j <= N + 1; ++j) {
                //for(int m = j/2; m <= j; ++m) 
                    //dp[i][j] = min(dp[i][j], max(dp[i - 1][j - m], dp[i][m]) + 1);
                /*
                int lb = 0, ub = j, lmd, rmd;
                while(lb <= ub) {
                    lmd = (lb + ub) >> 1, rmd = (lmd + ub) >> 1;
                    int lval = max(dp[i - 1][j - lmd], dp[i][lmd]);
                    int rval = max(dp[i - 1][j - rmd], dp[i][rmd]);
                    dp[i][j] = min(lval, rval) + 1;
                    if(lval == rval) lb = lmd + 1, ub = rmd - 1;
                    else if(lval < rval) ub = rmd;
                    else lb = lmd;
                }
                if(0 <= lb && lb <= j) dp[i][j] = min(dp[i][j], max(dp[i - 1][j - lb], dp[i][lb]) + 1);
                if(0 <= ub && ub <= j) dp[i][j] = min(dp[i][j], max(dp[i - 1][j - ub], dp[i][ub]) + 1);
                */
                int lb = 0, ub = j;
                while(lb <= ub) {
                    int md = (lb + ub) >> 1;
                    dp[i][j] = min(dp[i][j], max(dp[i - 1][j - md], dp[i][md]) + 1);
                    if(dp[i - 1][j - md] == dp[i][md]) lb = md + 1, ub = md - 1;
                    else if(dp[i - 1][j - md] > dp[i][md]) lb = md + 1;
                    else ub = md - 1;
                }
            }
        }
        return dp[K][N + 1];
    }
};

#ifdef __LOCAL_WONZY__
int main() {
    freopen("input.txt", "r", stdin);
    Solution solu;
    cout << solu.superEggDrop(1, 2) << endl;
    cout << solu.superEggDrop(2, 6) << endl;
    cout << solu.superEggDrop(3, 14) << endl;
    cout << solu.superEggDrop(3, 200) << endl;
    return 0;
}
#endif