poj 1724 ROADS 最短路

题目链接

n个节点， m条边， 一开始有K这么多的钱， 每条边有len， cost两个属性， 求1到n的最短距离， 花费要小于k。

dis数组开成二维的， dis[u][cost]表示到达u花费为cost的最短路径， 然后dij+堆优化。

路是单向的..

 #include <iostream>
 #include <vector>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <cmath>
 #include <map>
 #include <set>
 #include <string>
 #include <queue>
 using namespace std;
 #define pb(x) push_back(x)
 #define ll long long
 #define mk(x, y) make_pair(x, y)
 #define lson l, m, rt<<1
 #define mem(a) memset(a, 0, sizeof(a))
 #define rson m+1, r, rt<<1|1
 #define mem1(a) memset(a, -1, sizeof(a))
 #define mem2(a) memset(a, 0x3f, sizeof(a))
 #define rep(i, a, n) for(int i = a; i<n; i++)
 #define ull unsigned long long
 typedef pair<int, int> pll;
 const double PI = acos(-1.0);
 const double eps = 1e-;
 const int mod = 1e9+;
 const int inf = ;
 const int dir[][] = { {-, }, {, }, {, -}, {, } };
 const int maxn = 1e5+;
 int head[maxn], k, n, num, dis[][];
 struct node
 {
     int to, nextt, cost, d;
 }e[maxn*];
 void init() {
     num = ;
     mem1(head);
 }
 void add(int u, int v, int cost, int d) {
     e[num].to = v;
     e[num].nextt = head[u];
     e[num].cost = cost;
     e[num].d = d;
     head[u] = num++;
 }
 struct ee
 {
     int u, d, cost;
     bool operator < (ee a)const
     {
         if(d == a.d)
             return cost>a.cost;
         return d>a.d;
     }
     ee(){}
     ee(int u, int d, int cost):u(u), d(d), cost(cost){}
 };
 int dij() {
     priority_queue <ee> q;
     mem2(dis);
     dis[][] = ;
     q.push(ee(, , ));
     while(!q.empty()) {
         ee tmp = q.top(); q.pop();
         if(tmp.u == n)
             return tmp.d;
         for(int i = head[tmp.u]; ~i; i = e[i].nextt) {
             int v = e[i].to;
             if(dis[v][tmp.cost+e[i].cost]>tmp.d+e[i].d) {
                 if(tmp.cost+e[i].cost<=k) {
                     q.push(ee(v, tmp.d+e[i].d, tmp.cost+e[i].cost));
                     dis[v][tmp.cost+e[i].cost] = tmp.d+e[i].d;
                 }
             }
         }
     }
     return -;
 }
 int main()
 {
     int m, u, v, w, c;
     while(cin>>k) {
         cin>>n>>m;
         init();
         while(m--) {
             scanf("%d%d%d%d", &u, &v, &w, &c);
             add(u, v, c, w);
         }
         int ans = dij();
         printf("%d\n", ans);
     }
     return ;
 }