poj1062 昂贵的聘礼

思路：

最短路。枚举路径中节点的“等级”，合法才可以进行松弛操作。

实现：

#include <iostream>
#include <cstdio>
#include <vector>
#include <functional>
#include <queue>
using namespace std;

typedef pair<int, int> PP;
const int MAXN = 105, INF = 0x3f3f3f3f;

int M, N, P[MAXN], L[MAXN], X, dist[MAXN];
struct edge
{
    int to, cost;
};
vector<edge> G[MAXN];

int Dijkstra(int s, int l, int r)
{
    fill(dist + 1, dist + N + 1, INF);
    dist[s] = 0;
    priority_queue<PP, vector<PP>, greater<PP> > q;
    q.push(PP(0, s));
    while (!q.empty())
    {
        PP tmp = q.top(); q.pop();
        int id = tmp.second;
        if (tmp.first > dist[id]) continue;
        for (unsigned int i = 0; i < G[id].size(); i++)
        {
            edge & e = G[id][i];
            if (dist[e.to] >= dist[id] + e.cost && L[e.to] >= l && L[e.to] <= r)
            {
                dist[e.to] = dist[id] + e.cost;
                q.push(PP(dist[e.to], e.to));
            }
        }
    }
    int minn = INF;
    for (int i = 1; i <= N; i++)
        minn = min(minn, dist[i] + P[i]);
    return minn;
}

int main()
{
    cin >> M >> N;
    for (int i = 1; i <= N; i++)
    {
        cin >> P[i] >> L[i] >> X;
        int id, cost;
        for (int j = 0; j < X; j++)
        {
            cin >> id >> cost;
            G[i].push_back(edge{id, cost});
        }
    }
    int l = L[1], ans = INF;
    for (int i = max(l - M, 0); i <= l; i++)
    {
        int p = i, q = i + M;
        ans = min(ans, Dijkstra(1, p, q));
    }
    cout << ans << endl;
    return 0;
}