POJ 1724 ROADS（bfs最短路）

n个点m条边的有向图，每条边有距离跟花费两个参数，求1->n花费在K以内的最短路。

直接优先队列bfs暴力搞就行了，100*10000个状态而已。节点扩充的时候，dp[i][j]表示到达第i点花费为j时的最短路。没加优化16ms过，不知道discuss里面说bfs超时是怎么回事。。。。

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<fstream>
#include<sstream>
#include<bitset>
#include<vector>
#include<string>
#include<cstdio>
#include<cmath>
#include<stack>
#include<queue>
#include<stack>
#include<map>
#include<set>
#define FF(i, a, b) for(int i=a; i<b; i++)
#define FD(i, a, b) for(int i=a; i>=b; i--)
#define REP(i, n) for(int i=0; i<n; i++)
#define CLR(a, b) memset(a, b, sizeof(a))
#define debug puts("**debug**")
#define LL long long
#define PB push_back
#define eps 1e-10
using namespace std;

const int maxn = 111;
const int INF = 1e9;
int n, m, K, dp[maxn][10001];
struct Edge
{
    int from, to, dist, cost;
};
vector<int> G[maxn];
vector<Edge> edges;

void init()
{
    FF(i, 1, n+1)
    {
        G[i].clear();
        REP(j, K+1)
        {
            if(i == 1) dp[i][j] = 0;
            else dp[i][j] = INF;
        }
    }
    edges.clear();
}
void add(int u, int v, int d, int c)
{
    edges.PB((Edge){u, v, d, c});
    int nc = edges.size();
    G[u].PB(nc-1);
}

struct Node
{
    int u, d, c;
    bool operator < (const Node& rhs) const
    {
        return d > rhs.d;
    }
};

int bfs()
{
    priority_queue<Node> q;
    q.push((Node){1, 0, 0});
    while(!q.empty())
    {
        Node x = q.top(); q.pop();
        if(x.u == n) return x.d;
        int nc = G[x.u].size();
        REP(i, nc)
        {
            Edge e = edges[G[x.u][i]];
            if(e.cost + x.c > K) continue;
            if(dp[e.to][x.c+e.cost] > x.d+e.dist)
            {
                dp[e.to][x.c+e.cost] = x.d+e.dist;
                q.push((Node){e.to, x.d+e.dist, x.c+e.cost});
            }
        }
    }
    return -1;
}

int main()
{
    while(~scanf("%d%d%d", &K, &n, &m))
    {
        init();
        int a, b, c, d;
        REP(i, m)
        {
            scanf("%d%d%d%d", &a, &b, &c, &d);
            if(a != b) add(a, b, c, d);
        }
        printf("%d\n", bfs());
    }
    return 0;
}