POJ 3311 Hie with the Pie (状压DP)

题意：给一个起点和终点相同的图，一个矩阵表示各个点之间的距离，求经过所有的点，回到原点的最下路径，点可以重复走。

析：dp[s][i] 表示当前在 i 结点，还要遍历 s 的所有点并回到原点 0 的最短时间，状态转移方程也很简单。

dp[s][i] = min{dp[s|j][j] + d[i][j] }

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e16;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e6 + 10;
const int mod = 100000000;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}
int dp[1<<11][12];
int a[15][15];

int main(){
  while(scanf("%d", &n) == 1 && n){
    memset(dp, INF, sizeof dp);
    for(int i = 0; i <= n; ++i)
      for(int j = 0; j <= n; ++j)
        scanf("%d", &a[i][j]);
    int all = 1 << n+1;
    dp[all-1][0] = 0;
    for(int i = all-2; i >= 0; --i)
      for(int j = 0; j <= n; ++j)
        for(int k = 0; k <= n; ++k)
          dp[i][j] = min(dp[i][j], dp[i|(1<<k)][k] + a[j][k]);
    printf("%d\n", dp[0][0]);
  }
  return 0;
}