Poj_2926 Requirements

最新推荐文章于 2021-02-26 15:01:30 发布

最新推荐文章于 2021-02-26 15:01:30 发布 · 81 阅读

本文介绍了一个POJ平台上的编号为2926的模板题，该题涉及使用C++进行解决。代码中包含了完整的解决思路，从结构体定义到关键算法实现，并展示了如何通过循环和条件判断来处理问题。对于理解特定类型的算法题目非常有帮助。

不用说,完全的模板题,就是G++WA

#include <cstdio>
#include <ctime>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <string>
#include <set>
#include <stack>
#include <map>
#include <cmath>
#include <vector>
#include <iostream>
#include <algorithm>
#include <bitset>
#include <fstream>
using namespace std;


//LOOP
#define FF(i, a, b) for(int i = (a); i < (b); ++i)
#define FE(i, a, b) for(int i = (a); i <= (b); ++i)
#define FED(i, b, a) for(int i = (b); i>= (a); --i)
#define REP(i, N) for(int i = 0; i < (N); ++i)
#define CLR(A,value) memset(A,value,sizeof(A))
#define FC(it, c) for(__typeof((c).begin()) it = (c).begin(); it != (c).end(); it++)


//OTHER
#define SZ(V) (int)V.size()
#define PB push_back
#define MP make_pair
#define all(x) (x).begin(),(x).end()


//INPUT
#define RI(n) scanf("%d", &n)
#define RII(n, m) scanf("%d%d", &n, &m)
#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define RIV(n, m, k, p) scanf("%d%d%d%d", &n, &m, &k, &p)
#define RV(n, m, k, p, q) scanf("%d%d%d%d%d", &n, &m, &k, &p, &q)
#define RS(s) scanf("%s", s)


//OUTPUT
#define WI(n) printf("%d\n", n)
#define WS(n) printf("%s\n", n)


//debug
//#define online_judge
#ifndef online_judge
#define debugt(a) cout << (#a) << "=" << a << " ";
#define debugI(a) debugt(a) cout << endl
#define debugII(a, b) debugt(a) debugt(b) cout << endl
#define debugIII(a, b, c) debugt(a) debugt(b) debugt(c) cout << endl
#define debugIV(a, b, c, d) debugt(a) debugt(b) debugt(c) debugt(d) cout << endl
#else
#define debugI(v)
#define debugII(a, b)
#define debugIII(a, b, c)
#define debugIV(a, b, c, d)
#endif

typedef long long LL;
typedef unsigned long long ULL;
typedef vector <int> VI;
const int INF = 0x3f3f3f3f;
const double eps = 1e-10;
const int MOD = 100000007;
const int MAXN = 100010;
const double PI = acos(-1.0);

typedef double _int;
struct Node
{
    _int a[6], b[35];
} s[MAXN];
///结构体保存每个点的维数坐标(a),处理出来的状态(b)

struct Manhattan
{
    int k;
    _int Min[35], Max[35];
    void init()
    {
        REP(i, 1 << k)
        {
            Min[i] = 1e20, Max[i] = -1e20;
        }
    }

    ///work每次处理处所有的状态,add为增加一个点,del为删除一个点
    void work(int id)
    {
        REP(i, 1 << k)
        {
            _int sum = 0;
            REP(j, k)
            {
                if(i & (1 << j))
                    sum += s[id].a[j];
                else
                    sum -= s[id].a[j];
            }
            s[id].b[i] = sum;
        }
    }

    void add(int id)
    {
        work(id);
        REP(i, 1 << k)
        {
            if (Min[i] > s[id].b[i]) Min[i] = s[id].b[i];
            if (Max[i] < s[id].b[i]) Max[i] = s[id].b[i];
        }
    }

    _int get()
    {
        _int ret = -INF;
        REP(i, 1 << k)
        {
            _int t = Max[i] - Min[i];
            if(t > ret) ret = t;
        }
        return ret;
    }
} mandis;


int main()
{
    int n;
    mandis.k = 5;
    RI(n);
    mandis.init();
    FE(kase, 1, n)
    {
        REP(i, mandis.k)
            cin >> s[kase].a[i];
        mandis.add(kase);
    }
    printf("%.2lf\n", mandis.get());
    return 0;
}