ROADS

题目链接：http://poj.org/problem?id=1724

题目大意：

解题思路：

#include<iostream>
#include<fstream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<ctype.h>
#include<algorithm>
#include<string>
#define PI acos(-1.0)
#define maxn 1000
#define INF 1<<25
#define mem(a, b) memset(a, b, sizeof(a))
typedef long long ll;
using namespace std;
struct node
{
    int x, l, t;
    bool friend operator < (node s, node v) // 先根据路径长度非递减排序，如果长度相同根据费用非递减排序
    {
        if (s.l != v.l) return s.l > v.l;
        return s.t > v.t;
    }
};
struct E
{
    int d, l, t;
};
int k, n, r;
vector<E> v[110];
int bfs()
{
    node now, tmp;
    now.x = 1, now.l = 0, now.t = 0;
    priority_queue<node> q;
    q.push(now);
    while(!q.empty())
    {
        now = q.top();
        q.pop();
        if (now.x == n) return now.l;
        for (int u = 0; u < v[now.x].size(); u++) if (now.t + v[now.x][u].t <= k)
        {
            tmp.x = v[now.x][u].d, tmp.l = now.l + v[now.x][u].l, tmp.t = now.t + v[now.x][u].t;
            q.push(tmp);
        }
    }
    return -1;
}
int main ()
{
    scanf("%d%d%d", &k, &n, &r);
    while(r--)
    {
        int s;
        E e;
        scanf("%d%d%d%d", &s, &e.d, &e.l, &e.t);
        v[s].push_back(e);
    }
    printf("%d\n", bfs());
    return 0;
}

[POJ 1724] ROADS (有约束条件的最短路）

原文：http://blog.csdn.net/sio__five/article/details/18996155