POJ 3414 Paid Roads(状态压缩最短路)

题意：

求一条从1-N的最短路,当从点a到点b时有两个选择：

1.直接走过a, b, 消耗路费r

2.从a到b，当且仅当c也走过，在c处付路费min(p, r)

所以需要记录已经走过的顶点，判断是否需要在c处提前付路费

限制：

1 ≤ m, N ≤ 10,

0 ≤ P_i , R_i ≤ 100,

P_i ≤ R_i (1 ≤i ≤ m).

/*********************************************** * Author: fisty * Created Time: 2015-08-05 17:46:53 * File Name   : poj3411.cpp *********************************************** */#include <iostream>#include <cstring>#include <deque>#include <cmath>#include <queue>#include <stack>#include <list>#include <map>#include <set>#include <string>#include <vector>#include <cstdio>#include <bitset>#include <algorithm>using namespace std;#define Debug(x) cout << #x << " " << x <<endl#define Memset(x, a) memset(x, a, sizeof(x))const int INF = 0x3f3f3f3f;typedef long long LL;typedef pair<int, int> P;#define FOR(i, a, b) for(int i = a;i < b; i++)#define lson l, m, k<<1#define rson m+1, r, k<<1|1#define MAX_N 11int n, m;struct Rode{        int b, c, p, r;    Rode(int b, int c, int p, int r):b(b), c(c), p(p), r(r)    {}};struct State{    int u, cost, coverd;    State(int u, int cost, int coverd):u(u), cost(cost), coverd(coverd)    {}    bool operator < (const State& e1) const {        return e1.cost < cost;    }   };vector<Rode> G[MAX_N];priority_queue <State> que;int used[MAX_N][1<<MAX_N];void solve(){    Memset(used, 0);    que.push(State(1, 0, 1<<1)); //开始顶点为1，花费为0，走过顶点1    State result(-1,-1,-1);    while(que.size()){        State q = que.top(); que.pop();        if(used[q.u][q.coverd]) continue;        used[q.u][q.coverd] = 1;        if(q.u == n){            result = q;            break;        }        vector<Rode>::iterator it;        for(it = G[q.u].begin(); it != G[q.u].end(); ++it){            Rode e = *it;            int cost = q.cost;            if(q.coverd & (1 << e.c)){                cost += min(e.p, e.r);            }else{                cost += e.r;            }            que.push(State(e.b, cost , q.coverd|(1<<e.b)));        }    }    if(result.u == -1){        printf("impossible\n");    }else{        printf("%d\n", result.cost);    }}int main() {    //freopen("in.cpp", "r", stdin);    //cin.tie(0);    //ios::sync_with_stdio(false);    while(~scanf("%d%d", &n, &m)){        int a, b, c, p, r;        for(int i = 0;i < m; i++){            scanf("%d%d%d%d%d", &a, &b, &c, &p, &r);            G[a].push_back(Rode(b, c, p, r));        }        solve();    }    return 0;}