zoj3471 状压dp入门

Recently, researchers on Mars have discovered N powerful atoms. All of them are different. These atoms have some properties. When two of these atoms collide, one of them disappears and a lot of power is produced. Researchers know the way every two atoms perform when collided and the power every two atoms can produce.

You are to write a program to make it most powerful, which means that the sum of power produced during all the collides is maximal.

Input
There are multiple cases. The first line of each case has an integer N (2 <= N <= 10), which means there are N atoms: A1, A2, … , AN. Then N lines follow. There are N integers in each line. The j-th integer on the i-th line is the power produced when Ai and Aj collide with Aj gone. All integers are positive and not larger than 10000.

The last case is followed by a 0 in one line.

There will be no more than 500 cases including no more than 50 large cases that N is 10.

Output
Output the maximal power these N atoms can produce in a line for each case.

Sample Input
2
0 4
1 0
3
0 20 1
12 0 1
1 10 0
0

Sample Output
4
22

代码：

#include <iostream>#include <string>#include <cstring>#include <cstdio>#include <cmath>#include <cstdlib>#include <algorithm>#include <queue>#include <map>#include <vector>#define MST(s,q) memset(s,q,sizeof(s))#define INF 0x3f3f3f3f#define MAXN 1000005#define Lchild id<<1#define Rchild (id<<1)+1using namespace std;int dp[10000];int x[15][15];int DP(int n) {    MST(dp, 0);    for (int i = 0; i < (1 << n); i++) { // 二进制i中  1表示不存在，0表示存在        for (int j = 1; j <= n; j++) {            if (i & (1 << (j - 1))) continue; //  j已经消失            for (int k = 1; k <= n; k++) { // 用k撞j，k消失                if (j == k) continue;                if (i & (1 << (k - 1))) continue; // k不在状态i中存在                dp[i | (1 << (k - 1))] = max(dp[i | (1 << (k - 1))], dp[i] + x[j][k]);            }        }    }    int ans = 0;    for (int i = 1; i <= n; i++)        ans = max(ans, dp[(1 << n) - 1 - (1 << (i - 1))]);    return  ans;}int main() {    int N;    while (cin >> N && N) {        for (int i = 1; i <= N; i++)            for (int j = 1; j <= N; j++)                scanf("%d", &x[i][j]);        printf("%d\n", DP(N));    }}