题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1565
思路:跟上一篇 POJ 的状压dp很类似在这里就不过多阐述了
1 #include<iostream> 2 #include<algorithm> 3 #include<cstring> 4 #define M 20000 5 using namespace std; 6 typedef long long LL; 7 LL n, pass[M], now[22][M], mp[25][25]; 8 LL dp[22][M];//这里之所以不开1<<n 是因为虽然1<<n很大,但是其中有效状态比较小所以只开M个就足够了 9 bool check(int x, int y) { 10 return (pass[x] & pass[y]); 11 } 12 int main() 13 { 14 ios::sync_with_stdio(false); 15 while (cin >> n) { 16 if (n == 0) { 17 cout << "0" << endl; 18 continue; 19 } 20 memset(pass, 0, sizeof(pass)); 21 memset(now, 0, sizeof(now)); 22 memset(dp, 0, sizeof(dp)); 23 for (int i = 1; i <= n; i++) 24 for (int j = 1; j <= n; j++) 25 cin >> mp[i][j]; 26 int cnt = 0; 27 for (int i = 0; i < (1 << n); i++) 28 if (!(i&(i << 1)))//将可行状态保存下来 29 pass[cnt++] = i; 30 for (int i = 1; i <= n; i++) { 31 for (int j = 0; j < cnt; j++) { 32 int t = n; 33 for (int k = 1; k < (1 << n); k<<=1) { 34 if ((pass[j]&k)) 35 now[i][j] += mp[i][t];//now[i][j]代表第i行第j种状态的获取值 36 t--; 37 } 38 } 39 } 40 for (int i = 0; i < cnt; i++) 41 dp[1][i] = now[1][i]; 42 for (int i = 2; i <= n; i++) { 43 for (int j = 0; j < cnt; j++) { 44 for (int k = 0; k < cnt; k++) { 45 if (!check(j, k)) 46 dp[i][j] = max(dp[i][j], dp[i - 1][k] + now[i][j]); 47 } 48 } 49 } 50 LL ans = 0; 51 for (int i = 0; i < cnt; i++) 52 ans = max(dp[n][i], ans); 53 cout << ans << endl; 54 } 55 return 0; 56 }