poj3411 Paid Roads 状压+最短路

一个带权有向图，所有边<u,v>要么在v点收费，要么之前在某点交过费，求最小费用。

状压完图是带环的，求最短路即可。

#include <iostream>
#include <cstring>
#include <cctype>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <cstdio>
#include <set>
#include <map>
#include <queue>
#include <algorithm>
#define N 16
typedef long long LL;
#define fi first
#define se second
using namespace std;
struct edge
{
  int x , c , w1 , w2;
};
vector<edge> e[N];
int n , m , d[10][1 << 10];
bool f[10][1 << 10];
void work()
{
  int i , j , k , l , x; edge p;
  scanf("%d%d",&n,&m);
  while (m --)
  {
    scanf("%d%d%d%d%d",&j , &p.x , &p.c , &p.w1 , &p.w2);
    -- j , -- p.x , -- p.c;
    e[j].push_back(p);
  }
  for (i = 0 ; i < n ; ++ i)
    for (j = 0 ; j < 1 << n ; ++ j)
      d[i][j] = 1 << 30;
  memset(f , 0 , sizeof(f));
  queue< pair<int , int> > q;
  f[0][1] = 1 , q.push(make_pair(0 , 1)) ,d[0][1] = 0;
  while (!q.empty())
  {
    i = q.front().second , j = q.front().first , q.pop() , f[j][i] = 0;
    for (k = 0 ; k < e[j].size() ; ++ k)
    {
      x = e[j][k].x;
      if ((i & (1 << e[j][k].c)) && d[x][i | (1 << x)] > d[j][i] + e[j][k].w1)
      {
        d[x][i | (1 << x)] = d[j][i] + e[j][k].w1;
        if (!f[x][i | (1 << x)])
        {
          f[x][i | (1 << x)] = 1;
          q.push(make_pair(x , i | (1 << x)));
        }
      }
      if (d[x][i | (1 << x)] > d[j][i] + e[j][k].w2)
      {
        d[x][i | (1 << x)] = d[j][i] + e[j][k].w2;
        if (!f[x][i | (1 << x)])
        {
          f[x][i | (1 << x)] = 1;
          q.push(make_pair(x , i | (1 << x)));
        }
      }
    }
  }
  int ans = 1 << 30;
  for (i = 0 ; i < 1 << n ; ++ i)
    ans = min(ans , d[n - 1][i]);
  if (ans == 1 << 30)
    puts("impossible");
  else cout << ans << endl;
}

int main()
{
  work();
  return 0;
}