[USACO18DEC]Cowpatibility(容斥 or bitset优化暴力)

题面

题意：

给出ｎ个五元组(一个五元组的五个数互不相同)，我们称两个五元组不和谐，当且仅当任意元素都不相同，求有多少对五元组不和谐。

(Solution:)

很容易想到 Ans = 总共对数－和谐对数

而和谐对数包括５种：

• 一个数相同
• 二个数相同
• ...
• 五个数相同

所以我们就可以容斥了。

对于每次读进来的一组，我们计算它与之前读进来的和谐的个数。

于是我们就 (2^5)　枚举状态，然后 + (容斥系数) * (之前状态个数), 状态可以用map记因为状态有多个数, 所以用字符串作状态。

(Source)

#include <map>
#include <cmath>
#include <cctype>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <assert.h>
#include <algorithm>

using namespace std;

#define fir first
#define sec second
#define pb push_back
#define mp make_pair
#define LL long long
#define INF (0x3f3f3f3f)
#define mem(a, b) memset(a, b, sizeof (a))
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define Debug(x) cout << #x << " = " << x << endl
#define tralve(i, x) for (register int i = head[x]; i; i = nxt[i])
#define For(i, a, b) for (register int (i) = (a); (i) <= (b); ++ (i))
#define Forr(i, a, b) for (register int (i) = (a); (i) >= (b); -- (i))
#define file(s) freopen(s".in", "r", stdin), freopen(s".out", "w", stdout)
#define ____ debug("go
")

namespace io {
    static char buf[1<<21], *pos = buf, *end = buf;
    inline char getc()
    { return pos == end && (end = (pos = buf) + fread(buf, 1, 1<<21, stdin), pos == end) ? EOF : *pos ++; }
    inline int rint() {
        register int x = 0, f = 1;register char c;
        while (!isdigit(c = getc())) if (c == '-') f = -1;
        while (x = (x << 1) + (x << 3) + (c ^ 48), isdigit(c = getc()));
        return x * f;
    }
    inline LL rLL() {
        register LL x = 0, f = 1; register char c;
        while (!isdigit(c = getc())) if (c == '-') f = -1;
        while (x = (x << 1ll) + (x << 3ll) + (c ^ 48), isdigit(c = getc()));
        return x * f;
    }
    inline void rstr(char *str) {
        while (isspace(*str = getc()));
        while (!isspace(*++str = getc()))
            if (*str == EOF) break;
        *str = '';
    }
    template<typename T> 
        inline bool chkmin(T &x, T y) { return x > y ? (x = y, 1) : 0; }
    template<typename T>
        inline bool chkmax(T &x, T y) { return x < y ? (x = y, 1) : 0; }    
}

const int N = 4e5 + 1;

map<string, long long> M;
    
LL n;
int main() {
#ifndef ONLINE_JUDGE
    file("Cowpatibility");
#endif
    string a[6];
    
    cin >> n;
    LL ans = n * (n - 1) / 2;
    For (i, 1, n) {
        For (j, 1, 5) cin >> a[j];
        sort(a + 1, a + 6);
        LL res = 0;
        for (int s = 1; s < (1<<5); ++ s) {
            string str = ""; int ou = 0;
            for (int j = 1; j < 6; ++ j) {
                if (s & (1 << j - 1)) {
                    ou ++;
                    str += "," + a[j];
                }
            }
            if (ou & 1) res += M[str] ++;
            else res -= M[str] ++;
        }
        ans -= res;
    }
    cout << ans << endl;
}

然后因为bitset常数过于优秀, 总复杂度(O(frac{1}{32} imes n^2)), 虽然不是正解但也能过，而且跑的飞快。

#include <bitset>
#include <tr1/unordered_map>
#include <cctype>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <assert.h>
#include <algorithm>

using namespace std;

#define fir first
#define sec second
#define pb push_back
#define mp make_pair
#define LL long long
#define INF (0x3f3f3f3f)
#define mem(a, b) memset(a, b, sizeof (a))
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define Debug(x) cout << #x << " = " << x << endl
#define tralve(i, x) for (register int i = head[x]; i; i = nxt[i])
#define For(i, a, b) for (register int (i) = (a); (i) <= (b); ++ (i))
#define Forr(i, a, b) for (register int (i) = (a); (i) >= (b); -- (i))
#define file(s) freopen(s".in", "r", stdin), freopen(s".out", "w", stdout)
#define ____ debug("go
")

namespace io {
    static char buf[1<<21], *pos = buf, *end = buf;
    inline char getc()
    { return pos == end && (end = (pos = buf) + fread(buf, 1, 1<<21, stdin), pos == end) ? EOF : *pos ++; }
    inline int rint() {
        register int x = 0, f = 1;register char c;
        while (!isdigit(c = getc())) if (c == '-') f = -1;
        while (x = (x << 1) + (x << 3) + (c ^ 48), isdigit(c = getc()));
        return x * f;
    }
    inline LL rLL() {
        register LL x = 0, f = 1; register char c;
        while (!isdigit(c = getc())) if (c == '-') f = -1;
        while (x = (x << 1ll) + (x << 3ll) + (c ^ 48), isdigit(c = getc()));
        return x * f;
    }
    inline void rstr(char *str) {
        while (isspace(*str = getc()));
        while (!isspace(*++str = getc()))
            if (*str == EOF) break;
        *str = '';
    }
    template<typename T> 
        inline bool chkmin(T &x, T y) { return x > y ? (x = y, 1) : 0; }
    template<typename T>
        inline bool chkmax(T &x, T y) { return x < y ? (x = y, 1) : 0; }    
}using namespace io;

const int N = 5e4 + 1;

tr1::unordered_map<int, bitset<N> > buc;
    
int n, a[N][6];
int main() {
#ifndef ONLINE_JUDGE
    file("Cowpatibility");
#endif
    n = rint();
    For (i, 1, n) {
        For (j, 1, 5) 
            buc[ a[i][j] = rint() ].set(i);
    }
    bitset<N> tmp;
    int ans = 0;
    For (i, 1, n) {
        tmp.reset();
        For (j, 1, 5) 
            tmp |= buc[a[i][j]];
        ans += n - tmp.count();
    }
    cout << ans / 2 << endl;
}