【洛谷习题】售货员的难题
题目链接：https://www.luogu.org/problemnew/show/P1171
比较经典的状压DP吧，就是最后一个点不开O2卡不过去。。。
主要是对位运算熟，注意f[1][0]=0，其他状态为正无穷，别的看代码吧。
1 #include <cstdio> 2 #include <cstring> 3 4 using namespace std; 5 6 inline int get_num() { 7 int num = 0; 8 char c = getchar(); 9 while (c < '0' || c > '9') c = getchar(); 10 while (c >= '0' && c <= '9') 11 num = num * 10 + c - '0', c = getchar(); 12 return num; 13 } 14 15 const int maxn = 20; 16 17 int dist[maxn][maxn], f[1 << maxn][maxn]; 18 19 int main() { 20 int n, ans = 0x3f3f3f3f; 21 n = get_num(); 22 for (register int i = 0; i < n; ++i) 23 for (register int j = 0; j < n; ++j) 24 dist[i][j] = get_num(); 25 memset(f, 0x3f, sizeof(f)); 26 f[1][0] = 0; 27 for (register int i = 1; i < 1 << n; ++i) 28 for (register int j = 0; j < n; ++j) if (i >> j & 1) 29 for (register int k = 0; k < n; ++k) if ((i ^ 1 << j) >> k & 1) 30 if (f[i ^ 1 << j][k] + dist[k][j] < f[i][j]) 31 f[i][j] = f[i ^ 1 << j][k] + dist[k][j]; 32 for (register int i = 0; i < n; ++i) 33 if (f[(1 << n) - 1][i] + dist[i][0] < ans) 34 ans = f[(1 << n) - 1][i] + dist[i][0]; 35 printf("%d", ans); 36 return 0; 37 }
