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..
1 #include <iostream> 2 #include <vector> 3 #include <cstdio> 4 #include <cstring> 5 #include <algorithm> 6 #include <cmath> 7 #include <map> 8 #include <set> 9 #include <string> 10 #include <queue> 11 using namespace std; 12 #define pb(x) push_back(x) 13 #define ll long long 14 #define mk(x, y) make_pair(x, y) 15 #define lson l, m, rt<<1 16 #define mem(a) memset(a, 0, sizeof(a)) 17 #define rson m+1, r, rt<<1|1 18 #define mem1(a) memset(a, -1, sizeof(a)) 19 #define mem2(a) memset(a, 0x3f, sizeof(a)) 20 #define rep(i, a, n) for(int i = a; i<n; i++) 21 #define ull unsigned long long 22 typedef pair<int, int> pll; 23 const double PI = acos(-1.0); 24 const double eps = 1e-8; 25 const int mod = 1e9+7; 26 const int inf = 1061109567; 27 const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} }; 28 const int maxn = 1e5+5; 29 int head[maxn], k, n, num, dis[105][105]; 30 struct node 31 { 32 int to, nextt, cost, d; 33 }e[maxn*2]; 34 void init() { 35 num = 0; 36 mem1(head); 37 } 38 void add(int u, int v, int cost, int d) { 39 e[num].to = v; 40 e[num].nextt = head[u]; 41 e[num].cost = cost; 42 e[num].d = d; 43 head[u] = num++; 44 } 45 struct ee 46 { 47 int u, d, cost; 48 bool operator < (ee a)const 49 { 50 if(d == a.d) 51 return cost>a.cost; 52 return d>a.d; 53 } 54 ee(){} 55 ee(int u, int d, int cost):u(u), d(d), cost(cost){} 56 }; 57 int dij() { 58 priority_queue <ee> q; 59 mem2(dis); 60 dis[1][0] = 0; 61 q.push(ee(1, 0, 0)); 62 while(!q.empty()) { 63 ee tmp = q.top(); q.pop(); 64 if(tmp.u == n) 65 return tmp.d; 66 for(int i = head[tmp.u]; ~i; i = e[i].nextt) { 67 int v = e[i].to; 68 if(dis[v][tmp.cost+e[i].cost]>tmp.d+e[i].d) { 69 if(tmp.cost+e[i].cost<=k) { 70 q.push(ee(v, tmp.d+e[i].d, tmp.cost+e[i].cost)); 71 dis[v][tmp.cost+e[i].cost] = tmp.d+e[i].d; 72 } 73 } 74 } 75 } 76 return -1; 77 } 78 int main() 79 { 80 int m, u, v, w, c; 81 while(cin>>k) { 82 cin>>n>>m; 83 init(); 84 while(m--) { 85 scanf("%d%d%d%d", &u, &v, &w, &c); 86 add(u, v, c, w); 87 } 88 int ans = dij(); 89 printf("%d ", ans); 90 } 91 return 0; 92 }