poj 2531(按位暴力枚舉2^20很easy)

Network Saboteur

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 8492		Accepted: 3928

Description

A university network is composed of N computers. System administrators gathered information on the traffic between nodes, and carefully divided the network into two subnetworks in order to minimize traffic between parts.
A disgruntled computer science student Vasya, after being expelled from the university, decided to have his revenge. He hacked into the university network and decided to reassign computers to maximize the traffic between two subnetworks.
Unfortunately, he found that calculating such worst subdivision is one of those problems he, being a student, failed to solve. So he asks you, a more successful CS student, to help him.
The traffic data are given in the form of matrix C, where Cij is the amount of data sent between ith and jth nodes (Cij = Cji, Cii = 0). The goal is to divide the network nodes into the two disjointed subsets A and B so as to maximize the sum ∑Cij (i∈A,j∈B).

Input

The first line of input contains a number of nodes N (2 <= N <= 20). The following N lines, containing N space-separated integers each, represent the traffic matrix C (0 <= Cij <= 10000).
Output file must contain a single integer -- the maximum traffic between the subnetworks.

Output

Output must contain a single integer -- the maximum traffic between the subnetworks.

Sample Input

3
0 50 30
50 0 40
30 40 0

Sample Output

90

Source

Northeastern Europe 2002, Far-Eastern Subregion'

剛開始考慮用搜索之類的辦法，結果發現n<=20，直接2^20枚舉就好了。
貢獻了一次WA，需要注意的是 移位運算符<<運算優先級是低於減法運算符- 的。
因此 1<<n - 1 和 1<<(n-1)是等價的。
其實只需要枚舉[1, 1<<n-1] 就可以。
因爲後面的跟前面的是對稱的。
例如：
001   110
010   101
011   100

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define oo 0x3f3f3f3f
#define MAXN 25
using namespace std;
int data[30][30];
int n;
vector l;
vector r;
int cal(int num) {
    l.clear();
    r.clear();
    int i,j;
    for (i = 0; i < n; i++) {
        if (num & 1 > 0) l.push_back(i);
        else r.push_back(i);
        num >>= 1;
    }
    int sum = 0;
    for (i = 0; i < l.size(); i++) {
        for (j = 0; j < r.size(); j++) {
            sum += data[l[i]][r[j]];
        }
    }
    return sum;
}
int main()
{
    scanf("%d", &n);
    int i, j;
    for (i = 0; i < n; i++) {
        for (j = 0; j < n; j++) {
            scanf("%d", &data[i][j]);
        }
    }
    int high = 1 << (n - 1);//the - operator has more priority than << operator
    int imax = 0;
    for (i = 1; i <= high; i++) {
        j = cal(i);
        if (j > imax) imax = j;
        cout << j << endl;
    }
    //cout << imax << endl;
}