poj 3411 dij+状压

来源：互联网发布：windows rt刷win10移动编辑：程序博客网时间：2024/06/07 19:28

题意：

n个城市，编号为1~n，他们间有m条单向路，分别从a到b，可以在c处交P路费，也可以直接交R路费。

现在从1到n，问最少的花费是多少。

解析：

每次取边的时候判断一下。

代码：

#include <iostream>#include <cstdio>#include <cstdlib>#include <algorithm>#include <cstring>#include <cmath>#include <stack>#include <vector>#include <queue>#include <map>#include <climits>#include <cassert>#define LL long long#define lson lo, mi, rt << 1#define rson mi + 1, hi, rt << 1 | 1using namespace std;const int inf = 0x3f3f3f3f;const int maxn = 11;int n, m;struct Edge{    int a, b, c, P, R;    Edge() {}    Edge(int _a, int _b, int _c, int _P, int _R)    {        a = _a;        b = _b;        c = _c;        P = _P;        R = _R;    }};struct State{    int curPos, cost;    int cover;    State() {}    State(int _curPos, int _cost, int _cover)    {        curPos = _curPos;        cost = _cost;        cover = _cover;    }    bool operator < (const State& b)const    {        return cost > b.cost;    }};vector<Edge> edge[maxn];bool vis[maxn][1 << maxn];State ans;void dijkstra(){    priority_queue<State> q;    q.push(State(1, 0, 1 << 1));    memset(vis, false, sizeof(vis));    while (!q.empty())    {        State now = q.top();        q.pop();        if (vis[now.curPos][now.cover])            continue;        vis[now.curPos][now.cover] = true;        if (now.curPos == n)        {            ans = now;            break;        }        for (vector<Edge>::iterator it = edge[now.curPos].begin(); it != edge[now.curPos].end(); ++it)        {            Edge e = *it;            int cost = now.cost;            if (now.cover & (1 << e.c))            {                cost += min(e.P, e.R);            }            else            {                cost += e.R;            }            q.push(State(e.b, cost, now.cover | (1 << e.b)));        }    }}int main(){#ifdef LOCAL    freopen("in.txt", "r", stdin);#endif // LOCAL    while (~scanf("%d%d", &n, &m))    {        for (int i = 0; i < m; i++)        {            int a, b, c, P, R;            scanf("%d%d%d%d%d", &a, &b, &c, &P, &R);            edge[a].push_back(Edge(a, b, c, P, R));        }        ans.cost = -1;        ans.cover = -1;        ans.curPos = -1;        dijkstra();        if (ans.curPos == -1)        {            puts("impossible");        }        else        {            printf("%d\n", ans.cost);        }    }    return 0;}

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