UPC10582: Cowpatibility

时间限制: 1 Sec  内存限制: 128 MB
提交: 69  解决: 24
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题目描述

It turns out there is one factor that matters far more than any other when determining whether two cows are compatible as potential friends: whether they like similar flavors of ice cream!
Farmer John's N cows (2≤N≤50,000) have each listed their five favorite flavors of ice cream. To make this list concise, each possible flavor is represented by a positive integer ID at most 106. Two cows are compatible if their lists contain at least one common flavor of ice cream.

Please determine the number of pairs of cows that are NOT compatible

输入

The first line of input contains N. Each of the following N lines contain 5 integers (all different) representing the favorite ice cream flavors of one cow.

输出

Please output the number of pairs of cows that are not compatible.

样例输入

复制样例数据

4
1 2 3 4 5
1 2 3 10 8
10 9 8 7 6
50 60 70 80 90

样例输出

4

提示

Here, cow 4 is not compatible with any of cows 1, 2, or 3, and cows 1 and 3 are also not compatible.

来源/分类

USACO 2018 Dec, Gold 

 

每头牛的非空子集有31种情况，求出每头牛的所有子集，并记录一种子集被多少头牛所拥有

 

奇加偶减（奇数个类的计数和-偶数个类的计数和），记子集size为 1，2，3，4，5

1-2+3-4+5即为最终结果。

 

 

#include "bits/stdc++.h"


using namespace std;
typedef unsigned long long ull;
typedef long long ll;
const int maxn = 1e5;

int op[] = {-1, 1, -1, 1, -1, 1};

struct node {
    int n;
    int v[6];

} e[maxn];

bool operator<(const node &a, const node &b) {
    for (int j = 0; j < 5; j++) {
        if (a.v[j] < b.v[j]) return true;
        if (a.v[j] > b.v[j]) return false;
    }
    return false;
}

map<node, int> mp;

node ss(node &te, int j) {
    node ret = {0, {0, 0, 0, 0, 0, 0}};
    for (int i = 0; i < 5; i++) {
        if ((1 << i) & j) ret.v[ret.n++] = te.v[i];
    }
    return ret;
}

int main() {
    freopen("input.txt", "r", stdin);
    int n;
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        e[i].n = 5;
        for (int j = 0; j < 5; j++) {
            scanf("%d", &e[i].v[j]);
        }
        sort(e[i].v, e[i].v + 5);
        for (int j = 1; j < 32; j++) {
            mp[ss(e[i], j)]++;
        }
    }
    ll ans = 1ll * n * (n - 1) / 2;//运算时转为longlong，否则会溢出
    for (auto &p :mp) {
        ans -= 1ll * op[p.first.n] * p.second * (p.second - 1) / 2;
    }
    printf("%lld
", ans);
    return 0;
}