power oj 3136: Zed的影流部队Plus_c++一本通3136答案-优快云博客

本文分享了一个使用C++编写的程序案例，该程序通过动态规划等算法解决特定问题，并运用了宏定义、unordered_map等数据结构进行优化，旨在帮助读者理解如何分析并优化C++代码。

题目
要推的东西太多了，~~我不想写~~ ，你们要学会自己分析。

#include <bits/stdc++.h>
 
using namespace std;
 
#define rep(i, x, y) for(int i=(x);i<=(y);++i)
#define dep(i, x, y) for(int i=(x);i>=(y);--i)
const int INF = 0x3f3f3f3f;
const int maxn = 5e3;
int g[maxn][maxn];
int dp[maxn][1026];
int a[1000000];
int NUM[1000000];
 
int main() {
  //  freopen("01.in","r",stdin);
    int n, m;
    cin >> n >> m;
  //  n=n*m;
    for (int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
    }
 
    for (int i = 1; i <= m; i++) {
        for (int j = 0; i + j * m <= n; j++) {
            g[j + 1][i] = a[i + j * m];
            NUM[i]++;
        }
    }
    unordered_map<int, int> mp;
    unordered_map<int, int>::iterator it;
    int MIN = INF, Min = INF;
    for (int i = 1; i <= m; i++)
        rep(j, 0, 1024)dp[i][j] = INF;
    rep(j, 1, NUM[1]) {
        mp[g[j][1]]++;
    }
    for (int i = 0; i <= 1024; i++) {
        dp[1][i] = NUM[1] - mp[i];
        MIN = min(MIN, dp[1][i]);
    }
    Min = MIN;
    for (int i = 2; i <= m; i++) {
        mp.clear();
        MIN = INF;
        rep(j, 1, NUM[i]) {
            mp[g[j][i]]++;
        }
        n = NUM[i];
        for (int j = 0; j < 1024; j++) {
            for (it = mp.begin(); it != mp.end(); it++) {
                int X = it->first;
                int num = it->second;
                dp[i][j ^ X] = min(min(dp[i][j ^ X], dp[i - 1][j] + n - num), Min + n);
                MIN = min(dp[i][j ^ X], MIN);
            }
        }
        Min = MIN;
    }
    cout << dp[m][0] << endl;
}