POJ 3411 Paid Roads (状态压缩+BFS)

题意：有n座城市和m（1<=n，m<=10）条路。现在要从城市1到城市n。有些路是要收费的，从a城市到b城市，如果之前到过c城市，那么只要付P的钱，

如果没有去过就付R的钱。求的是最少要花多少钱。

析：BFS，然后由于走的路线不同，甚至边或者点都可能多走，所以用状态压缩。然后本题有坑啊，有重连，而且有很多条重边，所以多走几次就好了。

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 = 10 + 5;
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;
}

struct Node{
  int to, c, p, r, next;
  Node(){ }
  Node(int t, int cc, int pp, int rr, int n)
  : to(t), c(cc), p(pp), r(rr), next(n) { }
};
Node a[maxn*2];
int head[maxn], cnt;

void add_edge(int u, int v, int c, int p, int r){
  a[cnt] = Node(v, c, p, r, head[u]);
  head[u] = cnt++;
}

struct State{
  int state, pos;
  State(int s, int p) : state(s), pos(p) { }
};
int dp[1<<10][10];
int vis[10];

int bfs(){
  queue<State> q;
  int ans = INF;
  q.push(State(0, 0));
  memset(dp, INF, sizeof dp);
  memset(vis, 0, sizeof vis);
  dp[0][0] = 0;

  while(!q.empty()){
    State u = q.front();  q.pop();
    int state = u.state;
    if(u.pos + 1 == n){
      ans = min(ans, dp[state][u.pos]);
      continue;
    }
    for(int i = head[u.pos]; ~i; i = a[i].next){
      int v = a[i].to;
      int c = a[i].c;
      int r = a[i].r;
      int p = a[i].p;
      if(vis[v] > 50)  continue;
      ++vis[v];
      dp[state|(1<<v)][v] = min(dp[state|(1<<v)][v], dp[state][u.pos] + ((state&(1<<c)) ? p : r));
      q.push(State(state|(1<<v), v));
    }
  }
  return ans;
}

int main(){
  while(scanf("%d %d", &n, &m) == 2){
    cnt = 0;
    memset(head, -1, sizeof head);
    for(int i = 0; i < m; ++i){
      int u, v, c, p, r;
      scanf("%d %d %d %d %d", &u, &v, &c, &p, &r);
      add_edge(u-1, v-1, c-1, p, r);
    }
    int ans = bfs();
    if(ans == INF)  printf("impossible
");
    else  printf("%d
", ans);
  }
  return 0;
}