Poj 1724 ROADS (搜索 最短路 BFS优先队列)

题意：有n 城市，r条路，有k这么多的钱。每条路都有长度和花费两个参数，求从1到n最短且总花费不超过k的长度。

思路：优先队列。每次将长度最小的出队，然后判断花费，位超限就将后继点入队。

#include <cstdio>#include <cstring>#include <queue>using namespace std;int n,k,r,e;int head[105];struct Edges{    int v,c,l,next;}edge[10005];struct Node{    int u,cost,len;    Node (int _u,int _c, int _l)    {u=_u,cost=_c,len=_l;    }bool operator < (const Node& b) const {return len>b.len;}};void Addedge (int u, int v, int len, int cost){edge[e].next = head[u];edge[e].v = v;edge[e].l = len;edge[e].c = cost;head[u] = e++;}int Deal (){priority_queue<Node> q;q.push(Node(1,0,0));while (!q.empty()){Node temp = q.top();q.pop();if (temp.u == n)return temp.len;for (int i=head[temp.u];i != -1; i=edge[i].next)if (temp.cost + edge[i].c <= k)q.push (Node(edge[i].v,temp.cost+edge[i].c,temp.len+edge[i].l));}return -1;}int main (){scanf("%d%d%d",&k,&n,&r);e=0;    memset(head,-1,sizeof(head));for (int i=0; i<r;i++)    {int a,b,c,d;scanf("%d%d%d%d",&a,&b,&c,&d);Addedge(a,b,c,d);}printf("%d\n",Deal ());return 0;}